Counting sounds easy until the choices start piling up. Danish wants to make invitation cards in 5 different colours (red, blue, green, orange, yellow) and 3 different shapes (circle, square, rectangle). How many different cards can he make? You could draw every card — or you could notice that each colour can pair with each shape, so the total is 5 × 3 = 15 cards. The book sets this as a challenge: draw the tree diagram and count the 15 branches yourself.
Danish's father has a similar problem with a table: 4 materials (wood, plastic, glass, marble) and 3 shapes. The book draws the tree — 4 material branches, each splitting into 3 shape branches — and counts 12 choices. The shortcut is the same both times: multiply the number of options at each stage. The book states it formally: if event A can occur in m ways and event B in n ways, then the two events one after another can occur in m × n ways. The proof is a neat one — the pairs (ai, bj) are exactly the elements of the Cartesian product A × B, which has mn elements. And the rule extends: three events with m, n and k ways give m × n × k ways. 📌 Try yourself (book): three dice rolled together give 6 × 6 × 6 = 216 possible outcomes.
📌 History (book): Augustin Louis Cauchy (1789–1857) is called the father of permutation; Blaise Pascal and Pierre de Fermat (1607–1665) gave the idea of combinations of objects; Pascal and Leibniz are the founders of modern combinatorics.
Now seat 4 students on 4 chairs. The first chair has 4 options, the second only 3 (one student is seated), then 2, then 1 — so 4 × 3 × 2 × 1 = 24 ways. Products like this, of the first n natural numbers, appear so often that they get their own name: the factorial of n, written n! (or with an underline, n). So n! = n(n − 1)(n − 2)…3·2·1, and 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720. By definition 0! = 1 — a convention that makes all the later formulas work (the book justifies it again in §7.2: when n = r, ⁿPn must equal n!, which forces 0! = 1). A useful shortcut: n! = n(n − 1)!, so factorials cancel in fractions. The book's examples: 8!/6! = 8 × 7 = 56; 8·7·6·5 = 8!/4!; and 9!/(6!3!) = 84, worked three ways. 📌 Challenge (book): 8!/3! = 8 × 7 × 6 × 5 × 4 = 6720. 📌 History (book): the factorial notation (!) was introduced by Christian Kramp (1760–1826) in 1808.
گنتی اس وقت مشکل لگتی ہے جب انتخاب زیادہ ہو جائیں۔ دانش 5 رنگوں (سرخ، نیلا، سبز، نارنجی، پیلا) اور 3 شکلوں (دائرہ، مربع، مستطیل) سے دعوتی کارڈ بنانا چاہتا ہے۔ ہر رنگ ہر شکل کے ساتھ مل سکتا ہے، اس لیے کل کارڈ 5 × 3 = 15 ہوں گے۔ کتاب اسے چیلنج کے طور پر دیتی ہے: درختی خاکہ بنائیں اور 15 شاخیں خود گنیں۔
دانش کے والد کی میز کا مسئلہ بھی ایسا ہی ہے: 4 مواد (لکڑی، پلاسٹک، شیشہ، سنگ مرمر) اور 3 شکلیں — درختی خاکے میں 4 شاخیں، ہر ایک کی 3 شاخیں، یعنی کل 12 انتخاب۔ دونوں مثالوں سے ایک اصول نکلتا ہے: ہر مرحلے کے اختیارات کو ضرب دیں۔ کتاب اسے باقاعدہ بیان کرتی ہے: اگر واقعہ A، m طریقوں سے اور واقعہ B، n طریقوں سے ہو سکتا ہے تو دونوں یکے بعد دیگرے m × n طریقوں سے ہو سکتے ہیں۔ ثبوت یہ ہے کہ جوڑے (ai، bj) کارتیسی حاصل ضرب A × B کے ارکان ہیں، جن کی تعداد mn ہے۔ تین واقعات کے لیے m × n × k طریقے ہوں گے۔ 📌 خود آزمائیں (کتاب): تین پاسے ایک ساتھ پھینکنے پر 6 × 6 × 6 = 216 ممکنہ نتائج ہیں۔
اب 4 طلبہ کو 4 کرسیوں پر بٹھائیں: پہلی کرسی کے لیے 4، دوسری کے لیے 3، پھر 2، پھر 1 — یعنی 4 × 3 × 2 × 1 = 24 طریقے۔ پہلے n قدرتی اعداد کے اس حاصل ضرب کو فیکٹوریل کہتے ہیں، جسے n! لکھتے ہیں۔ پس n! = n(n − 1)(n − 2)…3·2·1، اور 1! = 1، 2! = 2، 3! = 6، 4! = 24، 5! = 120، 6! = 720۔ تعریف کے مطابق 0! = 1 ہے، تاکہ بعد کے تمام فارمولے درست رہیں۔ ایک کارآمد ترکیب: n! = n(n − 1)!، اس لیے کسروں میں فیکٹوریل آسانی سے کٹ جاتے ہیں۔ کتاب کی مثالیں: 8!/6! = 56؛ 8·7·6·5 = 8!/4!؛ اور 9!/(6!3!) = 84۔ 📌 چیلنج (کتاب): 8!/3! = 6720۔ 📌 تاریخ (کتاب): فیکٹوریل کی علامت (!) کرسچن کرامپ (1760–1826) نے 1808 میں متعارف کرائی۔
When the order matters, an arrangement is called a permutation. The number of permutations of r objects (r ≤ n) taken from n different objects is written ⁿPr or P(n, r), and the book proves the formula you will use all chapter: ⁿPr = n(n − 1)(n − 2)…(n − r + 1) = n!/(n − r)!. The proof is just the counting principle in action — the first place can be filled in n ways, the second in (n − 1) ways (no repetition allowed), the third in (n − 2), and so on until the rth place, which has n − r + 1 choices left. Multiply them, then tidy the product into factorials by multiplying top and bottom by (n − r)!.
The book's worked cases make the idea concrete. Three mathematics books for grades 1, 2 and 3, arranged in a row all at a time: ³P₃ = 3!/(3 − 3)! = 3!/0! = 6 ways (this is where 0! = 1 earns its keep). The letters of WORD taken all at a time: ⁴P₄ = 4! = 24 ways — and the book challenges you to draw all 24 pictorially. 📌 Do you know (book): in 1974 Ernő Rubik invented the Rubik's Cube, where every turn shows a permutation of the colours. Example 4: 4-digit numbers from the digits 1, 2, 3, 4, 5, 6 with no digit repeated: ⁶P₄ = 6 × 5 × 4 × 3 = 360.
Example 5 is the classic "books on a shelf" problem: 4 mathematics, 3 physics and 2 chemistry books, 9 spaces. (a) No restriction: 9! = 362,880 ways. (b) Books of the same subject together: tie each subject into one bundle — 3 bundles can be arranged in 3! ways, and inside the bundles the books arrange in 4!, 3! and 2! ways, giving 3! × 4! × 3! × 2! = 1728 ways. (c) Only the mathematics books together: the bundle plus the 5 single books make 6 objects, so 6! × 4! = 17,280 ways. 📌 Challenge (book): only the physics books together — by the same bundle trick, 3! × 7! = 30,240 ways. Example 6 seats 5 people on a bench: (a) no restriction, 5! = 120; (b) two particular people together — bundle them into one unit, 4! × 2! = 48; (c) the two not together — subtract from the total, 120 − 48 = 72 ways. That "total minus the unwanted case" move is one to remember.
جب ترتیب اہمیت رکھتی ہے تو اس ترتیب کو ترتیب (permutation) کہتے ہیں۔ n مختلف اشیاء میں سے r اشیاء کی ترتیبوں کی تعداد کو ⁿPr لکھتے ہیں، اور کتاب اس کا فارمولا ثابت کرتی ہے: ⁿPr = n(n − 1)(n − 2)…(n − r + 1) = n!/(n − r)!۔ ثبوت گنتی کے اصول کا سیدھا استعمال ہے — پہلی جگہ n طریقوں سے، دوسری (n − 1) طریقوں سے (تکرار کی اجازت نہیں)، تیسری (n − 2) طریقوں سے، اور اسی طرح r ویں جگہ تک جس کے لیے n − r + 1 انتخاب باقی ہیں۔ ان سب کو ضرب دیں تو فارمولا مل جاتا ہے۔
کتاب کی مثالیں اسے واضح کرتی ہیں۔ جماعت 1، 2 اور 3 کی تین ریاضی کی کتابیں ایک قطار میں: ³P₃ = 3!/0! = 6 طریقے (یہیں 0! = 1 کام آتا ہے)۔ لفظ WORD کے حروف سب ایک ساتھ: ⁴P₄ = 4! = 24 طریقے — اور کتاب چیلنج دیتی ہے کہ یہ 24 تصویری طور پر بنائیں۔ 📌 کیا آپ جانتے ہیں (کتاب): 1974 میں ارنو روبک نے روبک کیوب ایجاد کیا، جس کی ہر حرکت رنگوں کی ایک ترتیب ہے۔ مثال 4: اعداد 1 تا 6 سے بغیر تکرار کے 4 ہندسی اعداد: ⁶P₄ = 6 × 5 × 4 × 3 = 360۔
مثال 5 مشہور "شیلف پر کتابیں" کا مسئلہ ہے: 4 ریاضی، 3 طبیعیات اور 2 کیمیا کی کتابیں، 9 جگہیں۔ (الف) کوئی پابندی نہیں: 9! = 362,880 طریقے۔ (ب) ایک مضمون کی کتابیں ساتھ ساتھ: ہر مضمون کو ایک گٹھڑی بنائیں — 3 گٹھڑیاں 3! طریقوں سے، اور اندر کتابیں 4!، 3! اور 2! طریقوں سے، یعنی 3! × 4! × 3! × 2! = 1728 طریقے۔ (ج) صرف ریاضی کی کتابیں ساتھ: گٹھڑی اور 5 اکیلی کتابیں مل کر 6 اشیاء، یعنی 6! × 4! = 17,280 طریقے۔ 📌 چیلنج (کتاب): صرف طبیعیات کی کتابیں ساتھ ہوں تو اسی ترکیب سے 3! × 7! = 30,240 طریقے۔ مثال 6 میں 5 افراد بینچ پر: (الف) کوئی پابندی نہیں، 5! = 120؛ (ب) دو مخصوص افراد ساتھ — انہیں ایک اکائی بنائیں، 4! × 2! = 48؛ (ج) دونوں ساتھ نہ ہوں — کل سے ناپسندیدہ صورت گھٹائیں، 120 − 48 = 72 طریقے۔ "کل میں سے ناپسندیدہ گھٹانا" یہ ترکیب یاد رکھنے کی ہے۔
What if the objects are not all different? Take BITTER. If you first pretend the two T's are different — T₁ and T₂ — the 6 letters permute in 6! ways. But swapping T₁ and T₂ changes nothing you can see, so every visible arrangement has been counted 2! times. Divide: 6!/2! = 360 genuinely different arrangements. The general rule, given in the book's 📌 Remember box, is: if n objects include n₁ alike of one kind, n₂ alike of a second kind and n₃ alike of a third kind, the number of permutations taken all at a time is n!/(n₁!n₂!n₃!). Example 7 is the famous one: MISSISSIPPI has 11 letters, with I repeated 4 times, S 4 times, P 2 times and M once, so the count is 11!/(4!4!2!1!) = 34,650 ways.
Then the book turns the row into a circle. In a circular permutation, only the relative order around the circle matters — rotating everyone one seat along changes nothing. For three people A, B, C, the three linear arrangements A–B–C, B–C–A and C–A–B are all the same circular arrangement, so 3! linear permutations collapse to 3!/3 = 2! circular ones. In general, n objects around a circle give n!/n = (n − 1)! arrangements — Case-I, where clockwise and anticlockwise orders count as different. Example 8: 4 persons at a round table, clockwise different from anticlockwise: (4 − 1)! = 6 ways.
Case-II is for things you can flip over — a necklace, a ring of beads, a key ring — where a clockwise arrangement and its mirror image are the same thing. Then you divide once more by 2: (n − 1)!/2. Example 8 again: the same 4 persons, but now clockwise and anticlockwise identical: 3!/2 = 3 ways. 📌 Note (book): the two circular arrangements of A, B, C shown as reflections of each other count as one when clockwise and anticlockwise are considered identical. The trick for every round-table question is to ask first: can this arrangement be flipped? If yes, divide by 2; if no, stop at (n − 1)!.
اگر اشیاء سب مختلف نہ ہوں تو کیا ہوگا؟ لفظ BITTER لیں۔ اگر پہلے دونوں T کو مختلف سمجھیں — T₁ اور T₂ — تو 6 حروف 6! طریقوں سے ترتیب پاتے ہیں۔ لیکن T₁ اور T₂ کو بدلنے سے نظر آنے والی ترتیب نہیں بدلتی، اس لیے ہر نظر آنے والی ترتیب 2! مرتبہ گنی گئی ہے۔ تقسیم کریں: 6!/2! = 360 مختلف ترتیبیں۔ کتاب کے 📌 یاد رکھیں خانے کا عمومی اصول: اگر n اشیاء میں n₁ ایک قسم کی، n₂ دوسری قسم کی اور n₃ تیسری قسم کی مشابہ ہوں تو کل ترتیبیں n!/(n₁!n₂!n₃!) ہوں گی۔ مثال 7 مشہور ہے: MISSISSIPPI میں 11 حروف ہیں، I چار مرتبہ، S چار مرتبہ، P دو مرتبہ اور M ایک مرتبہ، اس لیے ترتیبیں 11!/(4!4!2!1!) = 34,650 ہیں۔
پھر کتاب قطار کو دائرہ بنا دیتی ہے۔ دائروی ترتیب میں صرف دائرے کے گرد نسبتی ترتیب اہمیت رکھتی ہے — سب کو ایک نشست آگے گھمانے سے کچھ نہیں بدلتا۔ تین افراد A، B، C کے لیے خطی ترتیبیں A–B–C، B–C–A اور C–A–B دائروی طور پر ایک ہی ہیں، اس لیے 3! خطی ترتیبیں 3!/3 = 2! دائروی ترتیبیں بن جاتی ہیں۔ عام طور پر، n اشیاء کو دائرے میں (n − 1)! طریقوں سے بٹھایا جا سکتا ہے — یہ صورت اول ہے، جہاں گھڑی وار اور مخالف سمت کی ترتیبیں مختلف گنی جاتی ہیں۔ مثال 8: 4 افراد گول میز پر، گھڑی وار مختلف ہو تو (4 − 1)! = 6 طریقے۔
صورت دوم ان چیزوں کے لیے ہے جنہیں الٹا جا سکتا ہے — ہار، موتیوں کی انگوٹھی، چابیوں کا حلقہ — جہاں گھڑی وار ترتیب اور اس کا عکس ایک ہی چیز ہے۔ تب ایک مرتبہ پھر 2 سے تقسیم کریں: (n − 1)!/2۔ مثال 8 دوبارہ: وہی 4 افراد، لیکن اب گھڑی وار اور مخالف سمت ایک جیسی ہوں تو 3!/2 = 3 طریقے۔ 📌 نوٹ (کتاب): A، B، C کی دو دائروی ترتیبیں جو ایک دوسرے کا عکس ہیں، ایک ہی گنی جاتی ہیں جب گھڑی وار اور مخالف سمت کو یکساں سمجھا جائے۔ ہر گول میز کے سوال میں پہلے یہ پوچھیں: کیا اس ترتیب کو الٹا جا سکتا ہے؟ اگر ہاں تو 2 سے تقسیم کریں، ورنہ (n − 1)! پر رک جائیں۔
A teacher picks Ahmad, Sana, Hamza and Danish for a writing-competition team. Writing the names as (Hamza, Ahmad, Danish, Sana) picks exactly the same team — the membership is what matters, not the order of listing. A combination of r objects taken from n objects is simply a subset of r objects from a set of n objects. Its count is written ⁿCr, C(n, r) or with the bracket notation, and the book proves it in one line of reasoning: every combination of r objects can be arranged among themselves in r! ways, so each combination gives rise to r! permutations — that is, ⁿCr × r! = ⁿPr. Since ⁿPr = n!/(n − r)!, dividing by r! gives ⁿCr = n!/(r!(n − r)!).
Two corollaries fall out at once. If r = n, then ⁿCn = n!/(n!0!) = 1 — there is exactly one way to take everything. If r = 0, then ⁿC₀ = n!/(0!n!) = 1 — exactly one way to take nothing. (Both lean on 0! = 1 again.) 📌 Remember (book): the formulae ⁿPr and ⁿCr are called counting formulae, because they count the possible number of ways without listing them all. That is the whole point of the chapter: you never have to write out 34,650 arrangements of MISSISSIPPI — the formula counts them for you. The one question to ask before choosing a formula is always the same: does the order matter? If yes, permute; if no, combine.
ایک استاد تحریری مقابلے کی ٹیم کے لیے احمد، ثنا، حمزہ اور دانش کو چنتا ہے۔ ناموں کو (حمزہ، احمد، دانش، ثنا) لکھنا بھی وہی ٹیم بناتا ہے — رکنیت اہمیت رکھتی ہے، فہرست کی ترتیب نہیں۔ n اشیاء میں سے r اشیاء کا انتخاب (combination) دراصل n اشیاء کے مجموعے کا r اشیاء پر مشتمل ذیلی مجموعہ ہے۔ اس کی تعداد کو ⁿCr لکھتے ہیں، اور کتاب اسے ایک سطر میں ثابت کرتی ہے: r اشیاء کا ہر انتخاب آپس میں r! طریقوں سے ترتیب پا سکتا ہے، اس لیے ہر انتخاب سے r! ترتیبیں بنتی ہیں — یعنی ⁿCr × r! = ⁿPr۔ چونکہ ⁿPr = n!/(n − r)! ہے، اس لیے r! سے تقسیم کرنے پر ⁿCr = n!/(r!(n − r)!) ملتا ہے۔
دو نتائج فوراً نکلتے ہیں۔ اگر r = n ہو تو ⁿCn = n!/(n!0!) = 1 — سب کچھ لینے کا صرف ایک طریقہ ہے۔ اگر r = 0 ہو تو ⁿC₀ = n!/(0!n!) = 1 — کچھ نہ لینے کا بھی صرف ایک طریقہ ہے۔ (دونوں پھر 0! = 1 پر انحصار کرتے ہیں۔) 📌 یاد رکھیں (کتاب): فارمولے ⁿPr اور ⁿCr کو گنتی کے فارمولے کہتے ہیں، کیونکہ یہ تمام طریقے لکھے بغیر ان کی تعداد گن لیتے ہیں۔ یہی اس باب کا اصل مقصد ہے: MISSISSIPPI کی 34,650 ترتیبیں لکھنے کی ضرورت نہیں — فارمولا آپ کے لیے گن لیتا ہے۔ فارمولا چننے سے پہلے ہمیشہ ایک ہی سوال پوچھیں: کیا ترتیب اہمیت رکھتی ہے؟ اگر ہاں تو ترتیب (permutation)، اگر نہیں تو انتخاب (combination)۔
Combinations show up wherever you choose without arranging. Example 9: Zain has 8 different fruits and wants 5 for a fruit chat — ⁸C₅ = 56 ways. Example 10: a class of 12 girls and 8 boys, choose 5 students — exactly 2 girls means ¹²C₂ × ⁸C₃ = 66 × 56 = 3696 ways; exactly 5 boys means ⁸C₅ = 56 ways; exactly 2 boys means ⁸C₂ × ¹²C₃ = 28 × 220 = 6160 ways. ⚠ Book slip: the book prints 36,960 for this last part, but its own working gives 28 × 220 = 6160 — the printed final number forgets the division by 3! in ¹²C₃, so 6160 is the correct value. 📌 Challenge (book): a restaurant offering 6 pizza flavours — choosing 2 flavours gives ⁶C₂ = 15 ways.
Complementary combinations are the chapter's best time-saver. Choosing r objects from n automatically leaves n − r behind, so every combination of r matches a combination of n − r: ⁿCr = ⁿCn−r. The book proves it both ways — by the pairing argument and by algebra, since (n − r)!((n − (n − r))!) = (n − r)!r!. 📌 Note (book): this is most useful when r > n/2 — for example, ¹²C₁₀ = ¹²C₂ = 66, a much friendlier calculation. Example 11 uses the same idea on geometry: a 6-sided figure has 6 vertices, joining any two gives ⁶C₂ = 15 line segments, and 6 of those are the sides themselves, leaving 15 − 6 = 9 diagonals.
The book then lines up permutation against combination in a table: order important versus order not important; arrangement versus selection; arranging coloured balls, alphabet letters or people on chairs versus selecting colours, team members or food items. The last three examples apply both ideas side by side. Example 12 (cryptography): a 6-character password from 26 letters and 10 digits — 36 characters in all. Without repetition it is a permutation, ³⁶P₆ = 1,402,410,240 passwords; with repetition allowed it is the counting principle, 36⁶. Example 13 (lottery odds): from 15 cards, choose the 5 that are the first five multiples of 2. If order matters there are ¹⁵P₅ = 360,360 possible selections, so the winning chance is 1/360,360 ≈ 0.000002775; if order does not matter there are only ¹⁵C₅ = 3003, so the chance is 1/3003 ≈ 0.00033 — the same win, 120 times likelier, just because order stopped mattering. Example 14: a DJ with 10 national songs choosing 5 — order matters, ¹⁰P₅ = 30,240 ways to play them; order does not matter, ¹⁰C₅ = 252 ways to select them.
انتخاب وہاں کام آتا ہے جہاں چنا جائے مگر ترتیب نہ دی جائے۔ مثال 9: زین کے پاس 8 مختلف پھل ہیں اور فروٹ چاٹ کے لیے 5 چاہیے — ⁸C₅ = 56 طریقے۔ مثال 10: 12 لڑکیوں اور 8 لڑکوں کی جماعت سے 5 طلبہ چنیں — ٹھیک 2 لڑکیاں یعنی ¹²C₂ × ⁸C₃ = 66 × 56 = 3696 طریقے؛ ٹھیک 5 لڑکے یعنی ⁸C₅ = 56 طریقے؛ ٹھیک 2 لڑکے یعنی ⁸C₂ × ¹²C₃ = 28 × 220 = 6160 طریقے۔ ⚠ کتاب کی غلطی: کتاب اس آخری حصے کے لیے 36,960 چھاپتی ہے، لیکن اس کے اپنے حساب سے 28 × 220 = 6160 بنتا ہے — چھپی ہوئی آخری عدد میں ¹²C₃ کی 3! سے تقسیم بھول گئی ہے، اس لیے درست جواب 6160 ہے۔ 📌 چیلنج (کتاب): 6 ذائقوں والا ریستوران — 2 ذائقے چننے کے ⁶C₂ = 15 طریقے۔
تکمیلی انتخاب اس باب کی بہترین وقت بچانے والی ترکیب ہے۔ n میں سے r اشیاء چننا خود بخود n − r کو پیچھے چھوڑ دیتا ہے، اس لیے r کے ہر انتخاب کے مقابلے میں n − r کا ایک انتخاب ہے: ⁿCr = ⁿCn−r۔ کتاب اسے دلیل اور الجبرا دونوں سے ثابت کرتی ہے۔ 📌 نوٹ (کتاب): یہ اس وقت سب سے کارآمد ہے جب r > n/2 ہو — مثلاً ¹²C₁₀ = ¹²C₂ = 66، جو کہیں آسان حساب ہے۔ مثال 11 اسی خیال کو ہندسہ پر لگاتی ہے: 6 رخی شکل کے 6 راس ہیں، کسی بھی دو کو ملانے سے ⁶C₂ = 15 قطعے بنتے ہیں، جن میں سے 6 خود شکل کے اضلاع ہیں، اس لیے قطر 15 − 6 = 9 ہیں۔
پھر کتاب ترتیب اور انتخاب کا موازنہ ایک جدول میں کرتی ہے: ترتیب اہم ہے بمقابلہ ترتیب اہم نہیں؛ ترتیب دینا بمقابلہ انتخاب کرنا؛ رنگین گیندیں، حروف تہجی یا کرسیوں پر لوگ ترتیب دینا بمقابلہ رنگ، ٹیم کے ارکان یا کھانے کی اشیاء چننا۔ آخری تین مثالیں دونوں خیالات کو ساتھ ساتھ لگاتی ہیں۔ مثال 12 (خفیہ نگاری): 26 حروف اور 10 ہندسوں سے 6 حرفی پاس ورڈ — بغیر تکرار کے یہ ترتیب ہے، ³⁶P₆ = 1,402,410,240 پاس ورڈ؛ تکرار کے ساتھ یہ گنتی کا اصول ہے، 36⁶۔ مثال 13 (لاٹری کے امکانات): 15 کارڈوں سے وہ 5 چنیں جو 2 کے پہلے پانچ اضعاف ہیں۔ اگر ترتیب اہمیت رکھتی ہے تو ¹⁵P₅ = 360,360 ممکنہ انتخاب ہیں، اس لیے جیتنے کا امکان 1/360,360 ≈ 0.000002775 ہے؛ اگر ترتیب اہمیت نہیں رکھتی تو صرف ¹⁵C₅ = 3003 انتخاب ہیں، اس لیے امکان 1/3003 ≈ 0.00033 ہے — وہی جیت، صرف ترتیب کے ختم ہونے سے 120 گنا زیادہ ممکن۔ مثال 14: 10 قومی نغموں والا ڈی جے 5 چنتا ہے — ترتیب اہمیت رکھتی ہے تو ¹⁰P₅ = 30,240 طریقے؛ نہیں رکھتی تو ¹⁰C₅ = 252 طریقے۔
Below are all questions of Exercises 7.1 to 7.4 from Unit 7 of the official book (printed pages 126, 129, 132–133, 138–139), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Where a step is derived rather than quoted from a book passage, or where the book's wording needs a note, the answer says so. Counting rule: one numbered question = one answer block; its parts (i), (ii), … are solved inside that block. That gives 7 + 15 + 14 + 20 = 56 blocks.
(i) 0! = 1, so 10!/(0!8!) = 10!/8! = 10 × 9 = 90 (everything from 8! downwards cancels). (ii) 12!/(3!9!) = (12 × 11 × 10)/(3 × 2 × 1) = 1320/6 = 220. (iii) 3!4! = 6 × 24 = 144, so 1440/144 = 10; and 5!2! = 120 × 2 = 240, so 2400/240 = 10. Total: 10 + 10 = 20. (iv) (n + 2)!/(n + 1)! = n + 2, since (n + 2)! = (n + 2)(n + 1)! and the (n + 1)! cancels: n + 2.
(i) Factor first: n³ − n = n(n² − 1) = n(n − 1)(n + 1) = (n − 1)n(n + 1) — three consecutive integers. As a factorial fraction: (n + 1)!/(n − 2)! = (n + 1)n(n − 1)(n − 2)!/(n − 2)! = (n + 1)n(n − 1). So n³ − n = (n + 1)!/(n − 2)! (for n ≥ 2). (ii) This product is the top r factors of n!: multiply top and bottom by (n − r)! to complete the factorial: n(n − 1)…(n − r + 1) = n!/(n − r)! — which is also ⁿPr.
Divide both sides by n!: (n + 4)(n + 3)(n + 2)(n + 1) = 3024. Now factor 3024 by trying consecutive integers: 9 × 8 × 7 × 6 = 72 × 42 = 3024. So n + 1 = 6, giving n = 5. Check: (5 + 4)! = 9! = 362,880, and 3024 × 5! = 3024 × 120 = 362,880. ✓
Multiply the whole equation by 9!: on the left, 9!/7! = 9 × 8 = 72 and 9!/8! = 9, so 72 + 9 = 81; on the right, x. Hence x = 81.
Split (2n + 1)! = 1 · 2 · 3 · … · (2n + 1) into its odd and even factors. The even factors are 2, 4, 6, …, 2n — there are n of them, and their product is 2 × 4 × … × 2n = 2ⁿ(1 × 2 × … × n) = 2ⁿn!. The odd factors are 1 · 3 · 5 · … · (2n + 1). So (2n + 1)! = [1 · 3 · 5 · … · (2n + 1)] × 2ⁿn!. Divide both sides by n!: (2n + 1)!/n! = [1 · 3 · 5 · … · (2n + 1)]2ⁿ, as required.
Take (n + 1)!/(r + 1)! common. The first term is (n + 1)!/(r + 1)! × (n + 2)/(r + 2), and the second term is (n + 1)!/(r + 1)! × 1. So the sum is (n + 1)!/(r + 1)! × [(n + 2)/(r + 2) + 1] = (n + 1)!/(r + 1)! × (n + r + 4)/(r + 2). Since (r + 1)!(r + 2) = (r + 2)!, the single fraction is (n + 1)!(n + r + 4)/(r + 2)!.
This is a derangement problem: arrange 4 balls so that none is in its matching box. Count with inclusion–exclusion. Total arrangements: 4! = 24. Subtract those with at least one ball in its own box, add back those with at least two, and so on: D₄ = 4!(1 − 1/1! + 1/2! − 1/3! + 1/4!) = 24(1 − 1 + 1/2 − 1/6 + 1/24) = 24 × (9/24) = 9 ways. (You can also list them: with balls 1, 2, 3, 4, the derangements are 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321 — nine in all.)
(i) ¹⁰P₅ = 10 × 9 × 8 × 7 × 6 = 30,240. (ii) ⁵P₂ = 5 × 4 = 20. (iii) ⁷P₇ = 7! = 5040. (iv) ¹⁰P₃ = 10 × 9 × 8 = 720.
(i) ⁿP₃ = n(n − 1)(n − 2) = 504. Three consecutive integers with product 504: 9 × 8 × 7 = 504, so n = 9. (ii) The right side is a product of 5 consecutive factors starting from 15, and ¹⁵Pn is a product of n consecutive factors starting from 15 — so n = 5. (iii) ⁿP₃ = n(n − 1)(n − 2) and ⁿ⁻¹P₂ = (n − 1)(n − 2). Their ratio is n(n − 1)(n − 2)/((n − 1)(n − 2)) = n. The given ratio is 540 : 1, so n = 540.
(i) By the formula, n · ⁿ⁻¹Pr−1 = n × (n − 1)!/((n − 1) − (r − 1))! = n(n − 1)!/(n − r)! = n!/(n − r)! = ⁿPr. In words: choose the first place in n ways, then fill the remaining r − 1 places from the remaining n − 1 objects. (ii) ⁿ⁻¹Pr + r · ⁿ⁻¹Pr−1 = (n − 1)!/(n − 1 − r)! + r(n − 1)!/(n − r)!. Write the first term with denominator (n − r)!: (n − 1)!(n − r)/(n − r)!. Adding: (n − 1)![(n − r) + r]/(n − r)! = (n − 1)! × n/(n − r)! = n!/(n − r)! = ⁿPr, as required. (In words: either a particular object is not used — ⁿ⁻¹Pr ways — or it is used, in which case it can sit in any of the r places and the rest fill in ⁿ⁻¹Pr−1 ways.)
All letters in each word are different, so it is just the factorial of the number of letters. (i) PYTHON has 6 letters: 6! = 720 words. (ii) NETWORK has 7 letters: 7! = 5040 words. (iii) COMPUTER has 8 letters: 8! = 40,320 words. (These "words" need not be meaningful — every arrangement counts.)
Order matters here — a red flag above a blue flag is a different signal from blue above red — so this is a permutation: ⁶P₂ = 6 × 5 = 30 signals.
"Any number" means 1, 2, 3, 4 or 5 flags at a time, and order matters within each signal. Add the cases: ⁵P₁ + ⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅ = 5 + 20 + 60 + 120 + 120 = 325 signals.
The odd digits available are 1, 3, 5, 7, 9 — five of them. Choose and arrange 4 of them in order: ⁵P₄ = 5 × 4 × 3 × 2 = 120 numbers.
Numbers between 100 and 1000 are 3-digit numbers. The hundreds digit cannot be 0, so it has 5 choices (1–5); the tens and units digits are then any 2 of the remaining 5 digits in order: ⁵P₂ = 20. Total: 5 × 20 = 100 numbers. For divisibility by 5, the units digit must be 0 or 5. Units digit 0: hundreds digit 5 choices, tens digit 4 remaining choices → 20 numbers. Units digit 5: hundreds digit 4 choices (not 0, not 5), tens digit 4 remaining choices → 16 numbers. Total divisible by 5: 20 + 16 = 36 numbers.
⚠ Reading stated: the book does not say how many digits the numbers have, so all lengths count — every 6-digit number formed from these digits is automatically greater than 35000, and so are the 5-digit numbers above 35000. Count both. 6-digit numbers: all 6 digits used: 6! = 720. 5-digit numbers greater than 35000: first digit 4, 5 or 6 (any of the remaining digits after it): 3 × ⁵P₄ = 3 × 120 = 360; first digit 3 with second digit 5 or 6 (so the number starts 35… or 36…): 2 × ⁴P₃ = 2 × 24 = 48. That is 360 + 48 = 408 five-digit numbers. Total: 720 + 408 = 1128 numbers. (If only 5-digit numbers are meant, the answer is 408.)
All five digits are used, so without any condition there are 5! = 120 numbers. (i) Tie 2 and 8 into one block: the block plus the other 3 digits make 4 objects, arranged in 4! ways, and inside the block 2 and 8 can sit as 28 or 82 (2 ways). Total: 4! × 2 = 48 numbers. (ii) Subtract from the total: 120 − 48 = 72 numbers with 2 and 8 not together.
For a 6-digit number the first digit cannot be 0: 5 choices (1–5), and the remaining 5 digits fill the other 5 places in 5! ways. Total: 5 × 120 = 600 numbers. Now fix 0 at the tens place (the second place from the right). The first digit still has 5 choices (1–5), and the remaining 4 places are filled by the remaining 4 digits in 4! = 24 ways. Total with 0 at the tens place: 5 × 24 = 120 numbers.
A multiple of 5 must end in 0 or 5. The only digit available here that works is 5, so the units digit is fixed as 5. The remaining 4 digits fill the first 4 places in 4! ways: 24 numbers.
Total arrangements without restriction: 8! = 40,320. Arrangements with the 2 English books together: tie them into one block — 7 objects in 7! ways, times 2! inside the block = 2 × 5040 = 10,080. Subtract: 40,320 − 10,080 = 30,240 ways with the English books never together.
Tie each subject into a block: the 2 blocks arrange in 2! ways; inside, the English books arrange in 3! ways and the Urdu books in 5! ways. Total: 2! × 3! × 5! = 2 × 6 × 120 = 1440 arrangements.
With 5 boys and 4 girls on 9 seats, alternation forces the pattern boy–girl–boy–girl–boy–girl–boy–girl–boy (it must start and end with a boy, since there is one more boy). The boys can occupy their 5 places in 5! = 120 ways, and the girls their 4 places in 4! = 24 ways. Total: 120 × 24 = 2880 ways.
(i) PAKISTAN has 8 letters with A repeated 2 times: 8!/2! = 40,320/2 = 20,160 arrangements. (ii) CURRICULUM has 10 letters: C twice, U three times, R twice (and I, L, M once each): 10!/(2!3!2!) = 3,628,800/24 = 151,200 arrangements. (iii) PROBABILITY has 11 letters: B twice, I twice (and P, R, O, A, L, T, Y once each): 11!/(2!2!) = 39,916,800/4 = 9,979,200 arrangements.
Fix B first. The remaining letters are A, N, A, N, A — five letters with A three times and N twice: 5!/(3!2!) = 120/12 = 10 permutations.
TRIGONOMETRY has 12 letters: T twice, R twice, O twice, and I, G, N, M, E, Y once each. Fix one T at the start and Y at the end. The 10 middle letters are T once, R twice, O twice, I, G, N, M, E once each: 10!/(2!2!) = 3,628,800/4 = 907,200 arrangements.
Nine marbles with repeats 4, 3 and 2: 9!/(4!3!2!) = 362,880/(24 × 6 × 2) = 362,880/288 = 1260 distinct arrangements.
Around a round table, rotations do not count as new arrangements, so use (n − 1)! (the book's standard Case-I reading, clockwise different from anticlockwise). (a) 8 persons: 7! = 5040 ways. (b) 7 persons: 6! = 720 ways. (c) 6 persons: 5! = 120 ways.
Seat the 5 men around the table first: (5 − 1)! = 24 ways. The 5 men leave 5 gaps between them, and the 5 women must occupy those gaps — one per gap, so that no two women sit together — in 5! = 120 ways. Total: 24 × 120 = 2880 ways.
Six digits with each of 7, 8 and 9 repeated twice: 6!/(2!2!2!) = 720/8 = 90 numbers. (Note that no number here starts with 0, so there is no leading-zero trap — every arrangement is a genuine 6-digit number.)
Choose the committees one after another: first committee of 3 from 15 (¹⁵C₃ = 455), then a committee of 5 from the remaining 12 (¹²C₅ = 792), then a committee of 4 from the remaining 7 (⁷C₄ = 35), then the last 3 form the final committee (³C₃ = 1). Multiply: 455 × 792 × 35 = 12,612,600 ways. Equivalently 15!/(3!5!4!3!) = 12,612,600. ⚠ Note: this counts the two 3-member committees as distinct groups (for example, formed in a stated order). If the two 3-member committees were treated as indistinguishable, the count would be halved to 6,306,300 — the book's wording does not say that, so 12,612,600 is the answer given here.
Tie the two particular D.C.Os into one block. The block plus the other 9 officers make 10 objects around the table: (10 − 1)! = 9! ways, and inside the block the two officers can sit in 2! = 2 orders. Total: 362,880 × 2 = 725,760 ways.
Fourteen officers around a round table (Case-I): (14 − 1)! = 13! = 6,227,020,800 ways.
The two tables are already distinguished by sex, so seat each group separately: the 9 males around their table in (9 − 1)! = 8! = 40,320 ways, and the 5 females around theirs in (5 − 1)! = 4! = 24 ways. Total: 40,320 × 24 = 967,680 ways.
With equal numbers, "no two of the same sex together" forces strict alternation around the circle. Seat the 5 men around the table: (5 − 1)! = 24 ways. The 5 men leave 5 gaps, and the 5 women fill them in 5! = 120 ways. Total: 24 × 120 = 2880 ways. (This is the same count as Q6 — there the "5 couples" wording hid the same men-and-women structure; here it is stated outright.)
A key ring can be flipped over, so a clockwise arrangement and its mirror image are the same — this is Case-II. The count is (8 − 1)!/2 = 5040/2 = 2520 ways.
A necklace can also be turned over, so clockwise and anticlockwise count as one — Case-II again: (10 − 1)!/2 = 362,880/2 = 181,440 necklaces.
(i) ⁵⁰C₅₀ = 1 (taking everything, one way). (ii) ¹⁰⁰⁰C₀ = 1 (taking nothing, one way). (iii) ¹⁰C₇ = ¹⁰C₃ = (10 × 9 × 8)/(3 × 2 × 1) = 120 — the complementary trick saves work here. (iv) ²⁰C₁₇ = ²⁰C₃ = (20 × 19 × 18)/6 = 1140.
(i) ³ⁿC₂ = 3n(3n − 1)/2 and ⁿC₂ = n(n − 1)/2, so their ratio is 3(3n − 1)/(n − 1). Set equal to 15: 3(3n − 1) = 15(n − 1), i.e. 9n − 3 = 15n − 15, i.e. 12 = 6n, so n = 2. Check: ⁶C₂ = 15 and ²C₂ = 1, ratio 15 : 1. ✓ (ii) ⁿPr/ⁿCr = r!, since ⁿPr = ⁿCr × r!. Here r! = 120/20 = 6, so r = 3.
(i) r! = ⁿPr/ⁿCr = 336/56 = 6, so r = 3. Then ⁿC₃ = n(n − 1)(n − 2)/6 = 56, i.e. n(n − 1)(n − 2) = 336 = 8 × 7 × 6, so n = 8, r = 3. (ii) ⚠ Flawed question — as printed, it has no solution. From the first ratio: ⁿ⁻¹Cr−1/ⁿCr = r/n (check: (n − 1)!/((r − 1)!(n − r)!) ÷ n!/(r!(n − r)!) = r/n). The ratio 1 : 3 gives r/n = 1/3, so n = 3r. From the second ratio: ⁿCr+1/ⁿCr = (n − r)/(r + 1) = 7/3. With n = 3r this becomes 2r/(r + 1) = 7/3, i.e. 6r = 7r + 7, i.e. r = −7 — impossible for a positive number of objects. So no positive integers n, r satisfy the printed question. (The likely intended question is ⁿCr−1 : ⁿCr : ⁿCr+1 = 1 : 3 : 7 — with the first term ⁿCr−1 instead of ⁿ⁻¹Cr−1 — which does have a clean solution: r/(n − r + 1) = 1/3 and (n − r)/(r + 1) = 7/3 give n = 19, r = 5. That answer belongs to the corrected question, not to the printed one.)
(i) ⁿCr + ⁿCr−1 = n!/(r!(n − r)!) + n!/((r − 1)!(n − r + 1)!). Put both over r!(n − r + 1)!: the first term becomes n!(n − r + 1)/(r!(n − r + 1)!) and the second becomes n!r/(r!(n − r + 1)!). Adding: n![(n − r + 1) + r]/(r!(n − r + 1)!) = n!(n + 1)/(r!(n − r + 1)!) = (n + 1)!/(r!((n + 1) − r)!) = ⁿ⁺¹Cr, as required. (This is Pascal's identity — the rule behind Pascal's triangle.) (ii) r · ⁿCr = r × n!/(r!(n − r)!) = n!/((r − 1)!(n − r)!). And (n − r + 1) · ⁿCr−1 = (n − r + 1) × n!/((r − 1)!(n − r + 1)!) = n!/((r − 1)!(n − r)!). The two sides are equal, as required.
Let the r consecutive integers be (m + 1), (m + 2), …, (m + r). Their product is (m + r)!/m!. Now consider the combination m+rCr = (m + r)!/(r!m!), which is a whole number (it counts combinations). Multiply both sides by r!: (m + r)!/m! = r! × m+rCr. So the product of the r consecutive integers is r! times an integer — that is, it is divisible by r!.
Order does not matter for a course programme — it is a selection: ⁸C₅ = ⁸C₃ = (8 × 7 × 6)/6 = 56 ways.
⚠ Reading stated: the book does not say how many vowels at a time, so this is read as arranging all five vowel letters A, E, I, O, U. Since order matters and all five letters are used, the count is 5! = 120 arrangements. (If "any number at a time" were meant, the count would be ⁵P₁ + ⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅ = 325, as in Exercise 7.2 Q6 — but that reading is not stated in the book here.)
Choose the Desi dishes and the Chinese dishes independently, then multiply: ⁶C₃ × ⁸C₂ = 20 × 28 = 560 ways.
Choose the 3 black cards from 26 and the 5 red cards from 26, then multiply: ²⁶C₃ × ²⁶C₅ = 2600 × 65,780 = 171,028,000 ways. (²⁶C₃ = (26 × 25 × 24)/6 = 2600; ²⁶C₅ = (26 × 25 × 24 × 23 × 22)/120 = 65,780.)
(i) ⁸C₃ × ⁷C₂ = 56 × 21 = 1176 ways. (ii) ⁸C₁ × ⁷C₄ = 8 × 35 = 280 ways. (iii) ⁸C₄ × ⁷C₁ = 70 × 7 = 490 ways. (iv) ⚠ Reading stated: five balls are being selected but there are 8 red balls, so "all the red balls" cannot literally mean all 8 — it is read as "all five balls selected are red": ⁸C₅ = 56 ways.
Joining any two of the 15 vertices gives ¹⁵C₂ = 105 line segments; 15 of those are the sides of the polygon, so the diagonals number 105 − 15 = 90. Joining any three vertices gives a triangle: ¹⁵C₃ = (15 × 14 × 13)/6 = 455 triangles.
For an n-sided polygon, diagonals = ⁿC₂ − n = n(n − 1)/2 − n = n(n − 3)/2. Set equal to 104: n(n − 3) = 208, i.e. n² − 3n − 208 = 0. Factor: (n − 16)(n + 13) = 0 (since 16 × 13 = 208 and 16 − 13 = 3). The positive root is n = 16 sides.
Without any restriction, 15 points give ¹⁵C₃ = 455 triangles. But any 3 of the 6 collinear points do not form a triangle — they form a straight line — so subtract ⁶C₃ = 20. Triangles formed: 455 − 20 = 435.
Choose the boys and the girls independently, then multiply: ¹⁰C₆ × ⁸C₃ = 210 × 56 = 11,760 ways. (¹⁰C₆ = ¹⁰C₄ = (10 × 9 × 8 × 7)/24 = 210; ⁸C₃ = (8 × 7 × 6)/6 = 56.)
The 2 particular persons are already in, so only 5 more members must be chosen — from the remaining 8 persons: ⁸C₅ = ⁸C₃ = (8 × 7 × 6)/6 = 56 committees.
Total selections: ¹⁷C₁₁ = ¹⁷C₆ = (17 × 16 × 15 × 14 × 13 × 12)/720 = 12,376 teams. Selections including a particular player: that player is already in, so choose the other 10 from the remaining 16: ¹⁶C₁₀ = ¹⁶C₆ = (16 × 15 × 14 × 13 × 12 × 11)/720 = 8008 teams.
(i) Exactly 3 women (and therefore 4 men): ⁸C₃ × ⁶C₄ = 56 × 15 = 840 committees. (ii) At most 3 women — add the cases 0, 1, 2 and 3 women. 0 women is impossible (only 6 men, but 7 seats). 1 woman: ⁸C₁ × ⁶C₆ = 8 × 1 = 8. 2 women: ⁸C₂ × ⁶C₅ = 28 × 6 = 168. 3 women: 840 (from part (i)). Total: 8 + 168 + 840 = 1016 committees. (iii) At least 5 women — add 5, 6 and 7 women. 5 women: ⁸C₅ × ⁶C₂ = 56 × 15 = 840. 6 women: ⁸C₆ × ⁶C₁ = 28 × 6 = 168. 7 women: ⁸C₇ = 8. Total: 840 + 168 + 8 = 1016 committees. (The two totals agree by symmetry: choosing 7 people with at most 3 women is the mirror image of choosing 7 with at least 5 women.)
Five questions across three sections, at least one from each: the only possible splits are (3, 1, 1) and (2, 2, 1). Split (3, 1, 1): choose which section gives all 3 questions (3 choices); that section contributes ³C₃ = 1 way, and each of the other two sections contributes ³C₁ = 3 ways: 3 × 1 × 3 × 3 = 27. Split (2, 2, 1): choose which section gives just 1 question (3 choices); that section contributes ³C₁ = 3 ways, and each of the other two contributes ³C₂ = 3 ways: 3 × 3 × 3 × 3 = 81. Total: 27 + 81 = 108 ways.
There are 6 possible letters (a–f) and 6 possible digits (0–5). First choose which 5 of the 8 positions hold letters: ⁸C₅ = 56 ways (the other 3 positions then hold digits). (a) With repetition allowed: each letter position has 6 choices and each digit position 6 choices: 6⁵ × 6³ = 7776 × 216 = 1,679,616 fillings per position-pattern. Total: 56 × 1,679,616 = 94,058,496 passwords. (b) Without repetition: the 5 letter positions must hold 5 different letters from 6, in order: ⁶P₅ = 720 ways; the 3 digit positions must hold 3 different digits from 6, in order: ⁶P₃ = 120 ways. Total: 56 × 720 × 120 = 4,838,400 passwords.
(i) Order important: arrange 3 songs from each list — ¹⁰P₃ from Teacher I's list and ¹⁰P₃ from Teacher II's list. ¹⁰P₃ = 720, so the total is 720 × 720 = 518,400 methods. (ii) Order not important: select 3 songs from each list — ¹⁰C₃ from each. ¹⁰C₃ = 120, so the total is 120 × 120 = 14,400 methods.
This chapter answers one question in many disguises: how many ways? The fundamental principle of counting is the root of everything — multiply the options at each stage. When the options shrink because nothing may repeat, the product becomes a factorial (n! = n(n − 1)…1, with 0! = 1). When order matters, you are permuting: ⁿPr = n!/(n − r)!. When order does not matter, you are combining: ⁿCr = n!/(r!(n − r)!) — each combination counted once instead of r! times. Repeated objects divide the factorial further (n!/(n₁!n₂!…)), circles divide by n ((n − 1)!), and flippable circles divide by 2 again ((n − 1)!/2). The complementary combination ⁿCr = ⁿCn−r halves your arithmetic whenever r is large. The chapter closes with the ideas at work: passwords, lottery odds, fruit chats, diagonals of polygons, committees and national songs.
Revision guidance: before touching any question, ask the two sorting questions — does order matter, and is repetition allowed? Those two answers pick the formula for you. Learn the bundle trick (tie together what must stay together, then multiply by the inside arrangements) and the subtraction trick (total minus the unwanted case) — together they solve most seating and shelf questions. For round tables, always ask whether the arrangement can be flipped; that single decision is worth a factor of 2. And in combination questions with two groups (boys and girls, Desi and Chinese dishes), choose from each group separately and multiply — never mix the groups into one big selection.
یہ باب کئی روپوں میں ایک ہی سوال کا جواب دیتا ہے: کتنے طریقے؟ گنتی کا بنیادی اصول ہر چیز کی جڑ ہے — ہر مرحلے کے اختیارات کو ضرب دیں۔ جب تکرار کی اجازت نہ ہونے سے اختیارات گھٹتے جائیں تو حاصل ضرب فیکٹوریل بن جاتا ہے (n! = n(n − 1)…1، اور 0! = 1)۔ جب ترتیب اہمیت رکھتی ہے تو ترتیب (permutation) ہے: ⁿPr = n!/(n − r)!۔ جب ترتیب اہمیت نہیں رکھتی تو انتخاب (combination) ہے: ⁿCr = n!/(r!(n − r)!) — ہر انتخاب ایک مرتبہ گنا جاتا ہے، r! مرتبہ نہیں۔ مشابہ اشیاء فیکٹوریل کو مزید تقسیم کرتی ہیں (n!/(n₁!n₂!…))، دائرے n سے تقسیم کرتے ہیں ((n − 1)!)، اور الٹے جا سکنے والے دائرے ایک مرتبہ پھر 2 سے ((n − 1)!/2)۔ تکمیلی انتخاب ⁿCr = ⁿCn−r جب r بڑا ہو تو حساب آدھا کر دیتا ہے۔ باب کے آخر میں یہ خیالات عملی کام میں دکھائے گئے ہیں: پاس ورڈ، لاٹری کے امکانات، فروٹ چاٹ، کثیرالاضلاع کے قطر، کمیٹیاں اور قومی نغمے۔
دہرائی کی رہنمائی: کسی بھی سوال کو چھونے سے پہلے دو ترتیب وار سوال پوچھیں — کیا ترتیب اہمیت رکھتی ہے، اور کیا تکرار کی اجازت ہے؟ ان دو جوابوں سے فارمولا خود چن لیا جاتا ہے۔ گٹھڑی کی ترکیب سیکھیں (جو ساتھ رہنا ہے اسے باندھیں، پھر اندرونی ترتیبوں سے ضرب دیں) اور گھٹانے کی ترکیب (کل میں سے ناپسندیدہ صورت گھٹائیں) — یہ دونوں مل کر نشست اور شیلف کے اکثر سوالات حل کر دیتی ہیں۔ گول میز کے لیے ہمیشہ پوچھیں کہ کیا ترتیب کو الٹا جا سکتا ہے؛ یہ ایک فیصلہ 2 کے عنصر کے برابر ہے۔ اور دو گروہوں والے انتخابی سوالوں میں (لڑکے اور لڑکیاں، دیسی اور چینی کھانے) ہر گروہ سے الگ الگ چنیں اور ضرب دیں — دونوں گروہوں کو ایک بڑے انتخاب میں کبھی نہ ملائیں۔
Topics covered: Counting principle and factorial · Permutations · Repeated and circular permutations · Combinations · Applications and complementary combinations · Exercises 7.1–7.4
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PunjabTest.com — Complete Guide Book • FSc Part 1 Mathematics, Chapter 7 (New Book, Session 2026–27) • English + Urdu