Combinations — nCr, Pascal's triangle: Question bank

2.7.11 · D5 · Maths › Statistics & Probability — Intermediate › Combinations — nCr, Pascal's triangle

Figure — Combinations — nCr, Pascal's triangle

Upar ka triangle dekho: har valid $(r n)$ dono orange edges ke andar baitha hai (jo sab $1$ 's hain, $r = 0$ aur $r = n$ cases). Jo bhi tum bahar likhoge un edges ke, usse $0$ maana jata hai — yahi visual meaning hai " $r > n$ gives $0$ " ki.

True ya false — justify karo

Har answer mein ek reason hona chahiye, sirf verdict nahi. "Kyunki" zor se bolo. Maano $0 \leq r \leq n$ (dono whole numbers) jab tak statement kuch aur na kahe.

(r n)

aur

^{n} P_{r}

same cheez count karte hain jab

r = 1

True — ek item choose karna aur ek item arrange karna identical hai, kyunki ek single item ki sirf

1! = 1

ordering hoti hai, to

(1 n) = \frac{^{n} P _{1}}{1 !} = n =^{n} P_{1}

(r n)

hamesha ek integer hota hai bhale hi formula ek fraction hai.

True — ye real committees count karta hai, to fractional nahi ho sakta; denominator mein

r! (n - r)!

hamesha

n!

ko exactly divide karta hai kyunki

n!

mein har group ki internal orderings ke factors shamil hain.

(r n) = (n - r n)

sirf tab hold karta hai jab

r < n - r

False — valid range

0 \leq r \leq n

ke har

r

ke liye hold karta hai;

r

items rakhne ka matlab hamesha

n - r

items discard karna bhi name karna hai, chahe kaunsa bada ho.

n

ko 1 badhane se hamesha

(r n)

badh jaata hai fixed

r

ke liye jab

1 \leq r \leq n

True — jab

r \geq 1

ho, to naye committees milte hain jo nayi person ko include karte hain aur saare purane bhi rehte hain, to count strictly badh jaata hai. (

r = 0

ke liye ye false hai:

(0 n) = 1

har

n

ke liye, to kuch nahi badalta.)

r

ko 1 badhane se hamesha

(r n)

badh jaata hai.

False — valid range

0 \leq r \leq n

mein row pehle upar jaati hai phir neeche;

(r n)

r = n /2

ke paas peak karta hai aur

r = n

par vapas

1

ho jaata hai, to middle se aage

r

badhane par count ghatta hai.

Pascal's triangle ki row

n

mein har entry ka sum

2^{n}

hota hai.

True —

n

items mein se har ek ya to subset mein hai ya nahi, jo

2^{n}

subsets deta hai total, jo exactly

\sum_{r = 0}^{n} (r n)

hai. Dekho Pascal's Triangle patterns.

(r n)

ordered pairs of teams count karta hai jab

n

logon ko

r

ka ek group aur

n - r

ka ek group mein split karo.

False — ye sirf ek unordered choice count karta hai; do groups mein split karna ek single decision hai, aur yahi wajah hai ki

(r n) = (n - r n)

(dono names same act describe karte hain).

Valid combination ke liye,

n

items distinct hone chahiye.

True — derivation

r!

se divide karta hai yeh maanke ki har internal arrangement genuinely different hai; agar items repeat hote hain, to kuch "arrangements" distinct nahi hote aur poore

r!

se divide karna over-correct kar deta hai.

Error pakdo

Har line mein ek galat claim ya step hai. Answer flaw ka naam deta hai aur use repair karta hai.

"7 mein se 3 ka committee

^{7} P_{3} = 210

choices deta hai."

Galat tool — permutations ordered selections count karte hain, lekin committee mein koi positions nahi hoti. 3 logon ki har ek committee

3! = 6

alag orders mein list ki ja sakti hai, to

^{7} P_{3}

har committee ko exactly

3!

baar count karta hai; us ordering ko mitane ke liye

3!

se divide karo, jo deta hai

(3 7) = 35

(0 n) = 0

kyunki kuch bhi choose na karna koi committee nahi deta."

Empty committee ek valid, single choice hai; kuch bhi na chunne ka exactly ek tarika hota hai, to

(0 n) = \frac{n !}{0 ! n !} = 1

0! = 1

use karke Factorials and 0! se.

"Pascal's triangle mein, har entry same row mein apne do neighbours ka sum hai."

Addition law uske diagonally upar ke do entries add karta hai, yaani previous row se

(r - 1 n - 1) + (r n - 1)

— same-row neighbours nahi.

(2 5) = \frac{5 !}{2 !} = 60

Denominator mein

(n - r)! = 3!

missing hai; sahi formula hai

\frac{5 !}{2 ! 3 !} = 10

—

(n - r)!

factor bhoolna sabse common slip hai.

(9 7)

undefined hai to hum ise

(- 2 7)

likhte hain."

Jab

r > n

ho to tum jo items hain unse zyada choose nahi kar sakte, to

(9 7) = 0

directly; negative

r

invent karne ki zaroorat nahi — ye sirf ek impossible selection hai.

"6 boys aur 4 girls (kul 10 log) ke club mein se 4-person team choose karo at least one girl ke saath: bas

(1 4) + (2 4) + (3 4) + (4 4)

add karo aur ho gaya."

Ye ignore karta hai ki baaki seats par 6 boys mein se kaun fill karega, to ye undercount karta hai; clean route hai total minus complement — sari teams minus all-boy teams —

(4 10) - (4 6) = 210 - 15 = 195

(dekho Probability — counting outcomes).

(98 100)

ko hundred-digit factorial chahiye, to ye haath se uncomputable hai."

Symmetry use karo:

(98 100) = (2 100) = \frac{100 \cdot 99}{2} = 4950

— multiply karne se pehle hamesha chote

r

mein convert karo.

"Kyunki order nahi maanta,

(2 4) = \frac{4 \cdot 3}{2} = 6

pairs hain, aur har pair 2 tareekon se arrange ho sakti hai, to actually 12 hain."

"Har pair ko 2 tareekon se arrange karo" step wahi ordering wapas laata hai jo hum ne remove ki thi;

2!

se divide karne ka poora point yahi tha ki un do arrangements ko ek maana jaye — answer

6

hi rehta hai.

Kyun wale questions

Reasoning hi answer hai — bare fact se kuch score nahi milta.

Hum

^{n} P_{r}

r!

se kyun divide karte hain aur kisi aur number se nahi?

Kyunki size

r

ka har unordered group apni internal orderings mein se ek baar aata hai, aur

r

items exactly

r!

tareekon se rearrange hoti hain;

r!

se divide karna un duplicates ko collapse karke ek single count mein le aata hai.

(r n)

symmetric kyun hai, yaani

r

choose karna

n - r

choose karne ke barabar kyun hai?

Jab bhi tum

r

jo rakhoge select karte ho, tumne saath hi saath uniquely

n - r

jo reject karoge use bhi select kar liya — ek action, do descriptions, to dono counts equal honee chahiye.

Pascal's rule upar ke do entries ko add kyun karta hai, multiply ya subtract nahi?

Kyunki ek special person ko fix karne se sare committees do disjoint, exhaustive cases mein split ho jaate hain (wo person andar, ya bahar); disjoint cases addition se combine hote hain, multiplication se kabhi nahi.

Hum sirf top

r

falling factors kyun likh sakte hain, jaise

7 \cdot 6 \cdot 5

, puri

7!

ki jagah?

r

factors hain hi

^{n} P_{r} = \frac{n !}{( n - r )!}

, kyunki

n!

(n - r)!

tail cancel ho jaati hai; poori factorial likhna sirf wo digits compute karne ki mehnat barbad karta hai jo immediately vanish ho jaati hain.

(a + b)^{n}

mein binomial coefficients exactly

(r n)

kyun nikalta hai?

Product expand karne ka matlab hai

n

factors mein se har ek mein ya to

a

b

choose karna; exactly

r

factors mein se

b

chunne ke tareekon ki sankhya

(r n)

hai, jo

a^{n - r} b^{r}

ka coefficient ban jaata hai — dekho Binomial Theorem.

Combination formula hold karne ke liye

n

items distinct kyun hone chahiye?

r!

se division ye maanke hota hai ki chosen group ki sari

r!

internal arrangements genuinely different hain; agar do items identical hain, to kuch "arrangements" coincide karte hain aur poore

r!

se divide karna bahut zyada remove kar deta hai.

(r n)

ek row ke middle mein sabse bada kyun hota hai, end par nahi?

Balanced subset chunne ke kahin zyada tarike hain compared to nearly-empty ya nearly-full ke — extremes

1

par pinned hain (sab andar ya sab bahar), jabki middle sizes mein bahut zyada freedom hai, isliye count centrally bulge karta hai.

Edge cases

Boundary inputs jahan formula phir bhi sahi behave karna chahiye. Test karo ki tum value aur kyun jaante ho.

(n n)

kya hai, aur kyun?

1

hai — sab

n

items choose karne ka exactly ek tarika hai (sab lo), aur formula deta hai

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

0! = 1

use karke.

(0 0)

kya hai?

1

— empty set mein exactly ek subset hota hai, empty set khud, aur

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

(r n)

kya hai jab

r > n

0

— tum set mein jo items hain unse zyada select nahi kar sakte, to aisa koi subset exist nahi karta; value genuinely zero hai, undefined nahi.

(0 5) + (5 5)

kya hai, aur ye kya represent karta hai?

1 + 1 = 2

— do sabse extreme choices, "kuch mat chuno" aur "sab chuno", dono exactly ek ek tarike se achievable hain; ye row 5 ke mirror ends hain.

Kya

(1 n) = n

counting meaning se obvious hai?

Haan —

n

distinct items mein se exactly ek item chunna

n

possibilities deta hai, ek per item, koi ordering nahi, aur formula confirm karta hai

\frac{n !}{1 ! ( n - 1 )!} = n

Kya Pascal's rule row ke edges par bhi kaam karta hai, jaise leading

1

milna?

Haan agar hum off-triangle entries ko

0

maanein:

(0 n) = (- 1 n - 1) + (0 n - 1) = 0 + 1 = 1

, to

1

's ka border naturally addition law se emerge karta hai.

(r n)

kya banta hai agar

r

negative ho?

Hum counting convention ke saath commit karte hain ki

(r n) = 0

har

r < 0

ke liye — ek subset mein negative number of elements nahi ho sakte, to aisi koi selection exist nahi karti aur uska count zero hai; ye choice Pascal's rule aur row-sum jaise identities ko boundaries par valid rakhti hai.

Recall Har trap ki ek-line summary

Sare traps teen questions tak reduce ho jaate hain: Kya order maayane rakhta hai? (permutation vs combination), Kya maine $(n - r)!$ rakha aur $0! = 1$ treat kiya? (formula boundaries), aur Kya mere cases disjoint hain? (kab add karein vs multiply). Ye teen seedhe rakho aur koi trap pakad nahi sakta.

Connections

Combinations — $\binom{n}{r}$ and Pascal's Triangle — parent note jise ye traps stress-test karte hain.
Permutations — nPr — "kya order maayane rakhta hai?" trap is boundary par rehta hai.
Factorials and 0! — $0! = 1$ convention empty-choice edge cases resolve karta hai.
Binomial Theorem — kyun coefficients hain hi combinations.
Probability — counting outcomes — "at least one" complement trick.
Pascal's Triangle patterns — row-sum $2^{n}$ aur addition-law direction traps.