Combinations — nCr, Pascal's triangle: Worked examples

2.7.11 · D3 · Maths › Statistics & Probability — Intermediate › Combinations — nCr, Pascal's triangle

Yeh page combinations ki workout room hai. Parent note ne machinery banayi; yahan hum use har tarah ki situation se guzarte hain jo ek problem de sakti hai — chhote edge cases, degenerate zeros, symmetry shortcuts, "at least one" complements, real-world word problems, aur ek exam-style trap.

Kuch bhi count karne se pehle, ek reminder seedhe shabdon mein: $(r n)$ (bolo " $n$ choose $r$ ") yaani $n$ chezon mein se $r$ chezon ke kitne alag groups nikaal sakte hain jab order matter nahi karta. Neeche har symbol usi idea ka ek alag roop hai.

The scenario matrix

Ek topic ko ek machine ki tarah socho jisme dials hain. Har dial setting ek alag tarah ki problem hai. Agar hum har setting test kar lein, toh exam mein koi bhi cheez humein surprise nahi kar sakti. Combinations ke liye yeh settings hain:

#	Case class	Kyun tricky hai	Example mein cover
A	Ordinary $0 < r < n$	seedhi counting, medium number expect karo	Ex 1
B	Edge: $r = 0$ (kuch bhi choose nahi)	$0$ kehne ka mann karta hai, sach yeh hai $1$	Ex 2
C	Edge: $r = n$ (sab kuch choose karo)	sirf ek hi full group exist karta hai	Ex 2
D	Degenerate: $r > n$ (zyada maang raha hai)	impossible — answer $0$ hona chahiye	Ex 2
E	Symmetry shortcut $r$ , $n$ ke karib	bada dikhta hai, $(r n) = (n - r n)$ se actually chhota nikalta hai	Ex 3
F	Pascal check — ek value do tarike se banao	addition law verify karta hai	Ex 4
G	"At least one" via complement	kai sub-cases simat jaate hain Total − None mein	Ex 5
H	Split selection (do groups mein se choose karo)	independent choices ko multiply karo	Ex 6
I	Real-world word problem	English → $n$ aur $r$ mein translate karo	Ex 7
J	Exam twist: permutation-vs-combination trap	order secretly matter karta hai (ya nahi karta)	Ex 8

Ab hum examples 1–8 se guzarte hain taaki upar ki har row cover ho jaye.

Example 1 — Cell A: the plain vanilla count

Worked example 9 players mein se 4-player team choose karo

9 distinct players mein se kitni alag 4-player teams ban sakti hain?

Forecast: padhne se pehle guess karo — kya yeh $100$ se zyada hoga ya kam? Ek number likh lo.

Step 1. Pehchano: ek team mein koi internal order nahi hota — do chosen players ko swap karne se wahi team milti hai. Yeh step kyun? Combinations, permutations nahi, exactly isliye kyunki order irrelevant hai. Agar humein parvah hoti ki kaun "pehle captain" hai, toh yeh ek permutation hota.

Step 2. Falling-factor form likho ( $9$ ke top $r = 4$ factors, $4!$ se divide karke): $(4 9) = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 !} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{24} .$ Yeh step kyun? Numerator $9 \cdot 8 \cdot 7 \cdot 6$ ordered count $^{9} P_{4}$ hai; $4! = 24$ se divide karne par woh ordering mit jaati hai jo hum nahi chahte (the "un-shuffle").

Step 3. Simplify karo: $\frac{3024}{24} = 126.$

Verify: Symmetry $(4 9) = (5 9)$ se cross-check karo; aur Pascal ki row $n = 9$ ke against: $1, 9, 36, 84, 126, 126, 84, 36, 9, 1$ — paanchwa entry sach mein $126$ hai. ✓

Example 2 — Cells B, C, D: teen edge/degenerate cases ek saath

Worked example

$(0 6)$ , $(6 6)$ , aur $(8 6)$ Teeno evaluate karo.

Forecast: inme se kaun sa $0$ ke barabar hoga?

Step 1 (Cell B, $r = 0$ ). "6 mein se kuch bhi choose karne ke kitne tarike hain?" Exactly ek hi tarika hai: empty set lo. $(0 6) = \frac{6 !}{0 ! 6 !} = \frac{6 !}{1 \cdot 6 !} = 1.$ Yeh step kyun? Yeh value $0!=1$ par depend karti hai. Agar tumne galti se socha $0! = 0$ toh zero se divide karte — nonsense. Genuinely ek empty group hota hai, isliye $1$ sahi hai.

Step 2 (Cell C, $r = n$ ). "6 mein se sab 6 choose karo." Sirf ek hi full group exist karta hai. $(6 6) = \frac{6 !}{6 ! 0 !} = 1.$ Yeh step kyun? Wahi $0! = 1$ engine — aur yeh Step 1 ke saath symmetry $(0 6) = (6 6)$ se match karta hai.

Step 3 (Cell D, $r > n$ ). "6 mein se 8 choose karo" impossible hai — tum jitne items hain unse zyada nahi nikal sakte. $(8 6) = 0.$ Yeh step kyun? Formula $\frac{6 !}{8 ! ( 6 - 8 )!}$ mein term $(6 - 8)! = (- 2)!$ undefined hai, jo nature ka tarika hai yeh chillane ka ki "impossible." Impossible chezon ki count $0$ hoti hai.

Verify: Poora $n = 6$ Pascal row hai $1, 6, 15, 20, 15, 6, 1$ — iske ends $1$ hain (Steps 1–2 se match karta hai), aur koi 9th entry nahi hai (Step 3 ke $0$ se match karta hai). ✓

Example 3 — Cell E: symmetry ek giant ko ant bana deta hai

Example 4 — Cell F: ek value do tarike se banao (Pascal's rule live)

Worked example

$(3 6)$ ko uske do Pascal parents se nikalo — ek picture ke saath Forecast: row $n = 5$ mein kaunse do numbers add hokar yeh dete hain?

Step 1. Pehle seedha compute karo: $(3 6) = \frac{6 \cdot 5 \cdot 4}{3 !} = \frac{120}{6} = 20.$ Yeh step kyun? Hume ek target chahiye taaki confirm kar sakein ki addition law isko reproduce karta hai.

Step 2. Pascal's rule apply karo $(r n) = (r - 1 n - 1) + (r n - 1)$ with $n = 6, r = 3$ : $(3 6) = (2 5) + (3 5) .$ Yeh step kyun? Story-proof (parent se): ek special person "Ravi" fix karo. Ya toh Ravi committee mein IN hai — phir baki $5$ mein se $r - 1 = 2$ aur choose karo, $(2 5)$ milte hain — ya Ravi OUT hai — baki $5$ mein se poore $3$ choose karo, $(3 5)$ milte hain. Yeh do cases overlap nahi karte aur sab kuch cover karte hain, isliye hum add karte hain. Figure mein red cell uske upar jhukte hue do black cells ka sum hai.

Step 3. Parents evaluate karo: $(2 5) = 10$ aur $(3 5) = 10$ , toh $10 + 10 = 20.$

Verify: $20 = 20$ ✓ — direct value Pascal sum ke barabar hai. Yahi addition law hai jo poora triangle grow karta hai.

Example 5 — Cell G: "at least one" via the complement

Worked example 5 physicists aur 3 chemists mein se,

at least one chemist ke saath 4 scientists choose karo Forecast: kya tum alag-alag "1 chemist" + "2 chemists" + "3 chemists" add karna pasand karoge, ya ek subtraction karna?

Step 1. $5 + 3 = 8$ scientists mein se 4 ke saare groups count karo: $(4 8) = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 !} = \frac{1680}{24} = 70.$ Yeh step kyun? "At least one chemist" messy hai (bahut cases), lekin iska opposite — "bilkul koi chemist nahi" — ek single clean count hai. Poori universe se start karo.

Step 2. Complement count karo: zero chemists wale groups, yani saare 4, 5 physicists mein se choose kiye: $(4 5) = (1 5) = 5.$ Yeh step kyun? Symmetry $(4 5) = (1 5)$ use ki taaki yeh instant ho jaye. Yeh "none" case hai jo hum hatayenge.

Step 3. Subtract karo: $70 - 5 = 65.$ Yeh step kyun? Total groups minus jinhe hum nahi chahte (no chemist) = exactly "at least one chemist" wale groups.

Verify (long way, agree karna chahiye): exact number of chemists ke hisaab se count karo:

1 chemist: $(1 3) (3 5) = 3 \cdot 10 = 30$
2 chemists: $(2 3) (2 5) = 3 \cdot 10 = 30$
3 chemists: $(3 3) (1 5) = 1 \cdot 5 = 5$

Sum $= 30 + 30 + 5 = 65.$ ✓ Complement method se match karta hai.

Example 6 — Cell H: split selection (independent choices ko multiply karo)

Worked example 5 boys aur 6 girls mein se 2 boys AUR 3 girls ki committee

Forecast: kya hum dono counts ko add karenge ya multiply?

Step 1. Boys choose karo, girls ko ignore karte hue: $(2 5) = \frac{5 \cdot 4}{2} = 10.$ Yeh step kyun? Boy-choice ek self-contained combination hai; boys mein order irrelevant hai.

Step 2. Girls independently choose karo: $(3 6) = \frac{6 \cdot 5 \cdot 4}{6} = 20.$ Yeh step kyun? Girls ke liye same logic, aur yeh choice is baat se interact nahi karti ki kaunse boys pick hue.

Step 3. Multiplication principle se combine karo: $(2 5) \cdot (3 6) = 10 \cdot 20 = 200.$ Yeh step kyun? 10 boy-groups mein se har ek ke liye, saare 20 girl-groups possible hain, toh total pairs $= 10 \times 20$ . Hum multiply karte hain (independent stages), unlike Example 5 jahan disjoint cases add kiye gaye the.

Verify: Units sanity — hume committees chahiye thi, aur $10$ (boy-ways) $\times$ $20$ (girl-ways) $= 200$ full committees. Har committee mein exactly 2 boys aur 3 girls hain jaise required, koi double-counting nahi. ✓

Example 7 — Cell I: ek real-world word problem

Worked example Pizza toppings

Ek shop 8 distinct toppings offer karta hai. Ek "supreme" pizza unme se exactly 3 use karta hai, aur toppings repeat nahi ho sakti. Kitne alag supreme pizzas exist karte hain? Kya pehle pepperoni daalna ya baad mein daalna matter karta hai?

Forecast: guess karo ki yeh $60$ se kam hoga ya zyada.

Step 1. English → numbers mein translate karo: "8 distinct toppings" ⇒ $n = 8$ ; "exactly 3, no repeats" ⇒ $r = 3$ ; "ek pizza ko order yaad nahi rehta jisme toppings dale" ⇒ order irrelevant ⇒ combination. Yeh step kyun? "Order doesn't matter" phrase hi $(3 8)$ aur $^{8} P_{3}$ mein poora decision hai. {ham, olive, chili} wala pizza {chili, olive, ham} wale pizza se identical hai.

Step 2. Compute karo: $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 !} = \frac{336}{6} = 56.$ Yeh step kyun? Top teen falling factors $3!$ se divide — standard un-shuffle.

Verify: Symmetry cross-check $(3 8) = (5 8)$ ; aur Pascal's rule $(3 8) = (2 7) + (3 7) = 21 + 35 = 56.$ ✓ Do independent checks agree karte hain.

Example 8 — Cell J: the exam trap (order secretly matters)

Worked example Same numbers, alag question — switch pakdo

8 runners mein se, (a) final mein advance karne ke liye 3 ka group choose karne ke kitne tarike hain, aur (b) unme se 3 ko gold, silver, bronze medals award karne ke kitne tarike hain?

Forecast: kya (a) aur (b) equal honge? Agar nahi, toh kaun sa bada hoga aur kitne factor se?

Step 1 (a) — group, no order. Advance karna ek plain committee hai: $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 !} = 56.$ Yeh step kyun? Chahe runner X "1st-named" advance kare ya "3rd-named" — koi fark nahi padta — wahi trio advance karta hai. Order irrelevant ⇒ combination.

Step 2 (b) — medals, order matters. Gold ≠ silver ≠ bronze, isliye unhi teen runners ki arrangement ab alag outcomes deti hai. Yeh ek permutation hai: $^{8} P_{3} = 8 \cdot 7 \cdot 6 = 336.$ Yeh step kyun? Yahan kaun gold leta hai aur kaun bronze leta hai swap karna genuinely ek alag result hai — isliye hum $3!$ se divide nahi karenge. Yahi parent note ke "Error A" ka trap hai, ulta chalaaya gaya.

Step 3 — relationship dekho. Dono answers exactly un-shuffle factor se differ karte hain: $^{8} P_{3} = (3 8) \cdot 3! = 56 \cdot 6 = 336.$ Yeh step kyun? Har unordered trio (56 hai) ko medal-order mein $3! = 6$ tarike se arrange kiya ja sakta hai, $336$ milta hai. Yahi master relation $(r n) = \frac{^{n} P _{r}}{r !}$ hai, aage ki taraf padha gaya.

Verify: $56 \cdot 6 = 336$ ✓, aur $336/6 = 56$ ✓. Single test — "kya do chosen logon ko swap karne se answer change hota hai?" — cleanly (a) aur (b) ko alag karta hai.

Active recall

(0 6) = 1

kyun hota hai,

0

kyun nahi?

Kuch bhi choose karne ka exactly ek hi tarika hai (empty set); formula deta hai

\frac{6 !}{0 ! 6 !} = 1

using

0! = 1

(8 6)

kya hai aur kyun?

0

— tum jitne items exist karte hain unse zyada choose nahi kar sakte;

(6 - 8)!

undefined hai, impossibility signal karta hai.

(48 50)

jaldi compute karne ka tarika?

Symmetry use karo

(48 50) = (2 50) = \frac{50 \cdot 49}{2} = 1225

Ek pool mein se "at least one chemist" — strategy kya hai?

Total minus complement:

(4 8) - (4 5) = 70 - 5 = 65

2 boys aur 3 girls — sub-counts add karo ya multiply?

Multiply karo (independent stages):

(2 5) (3 6) = 10 \cdot 20 = 200

8 mein se group of 3 vs gold/silver/bronze — kaun sa bada hai aur kitna?

Medals factor

3!

se bade hain:

^{8} P_{3} = 336 = 6 \cdot (3 8) = 6 \cdot 56

Connections

Combinations — nCr, Pascal's triangle — parent machinery jinhe yeh examples exercise karti hain.
Permutations — nPr — Example 8 ka trap order-matters side par rehta hai.
Factorials and 0! — Example 2 ke edge cases $0! = 1$ par hinge karte hain.
Binomial Theorem — wahi $(r n)$ $(a + b)^{n}$ ke coefficients ke roop mein appear karte hain.
Probability — counting outcomes — yeh counts probabilities ke numerators/denominators bante hain.
Pascal's Triangle patterns — Example 4 ka addition law triangle ka growth rule hai.