2.7.10
Permutation ek ordered positions ek-ek karke fill karte ho , aur har position par sochte ho kitne choices bache hain . Un choices ko multiply karo — bas yahi poora subject hai. Restrictions sirf ye change karti hain ki kisi step par kitne choices hain .
Objects ka ek definite order mein arrangement. n distinct objects mein se r == n P r ==
WHY order matters: A B B A alag arrangements hain. Agar order matter nahi karta (jaise ek team chunna), to hum combinations karte, permutations nahi.
n r
Position
Bache hue choices
1st
n
2nd
n − 1
3rd
n − 2
… …
r n − ( r − 1 ) = n − r + 1
Multiplication principle se, choices ko multiply karo:
n P r = n ( n − 1 ) ( n − 2 ) ⋯ ( n − r + 1 )
Ye step kyun? Har position ek independent decision hai pichle walo ko dete hue , aur independent-aur-phir decisions multiply hote hain.
Ab ise compact banao. "Leftover" factorial ( n − r )!
n P r = ( n − r )! n ( n − 1 ) ⋯ ( n − r + 1 ) ⋅ ( n − r )! = ( n − r )! n !
Exam ke almost har sawaal mein in paanch patterns mein se ek hota hai. Inhe master karo aur 80% marks cover ho jaate hain.
WHY: Sabse tight constraint ko pehle handle karo, warna baad mein freedom khatam ho jaayegi.
Us block ko ek super-object maano, arrange karo, phir block ke andar arrange karo.
Pehle baaki objects arrange karo, phir "must-be-apart" wale objects ko unke beech ke gaps mein daalo.
Forced objects rakh do, phir baaki ko permute karo.
Total count karo, forbidden subtract karo.
Worked example E1 — Plainn P r { A , B , C , D , E } 5 P 3 = 2 ! 5 ! = 5 ⋅ 4 ⋅ 3 = 60 5 ⋅ 4 ⋅ 3
Worked example E2 — Fixed position (Pattern 1){ A , B , C , D , E } vowel se shuru ho .
Step 1: pehla letter vowel ho → 2 A E Pehle kyun? Ye constraint hai. Step 2: baaki 4 letters ko 4 slots mein arrange karo → 4 ! = 24 2 × 24 = 48
Worked example E3 — Together (Pattern 2, glue)Kitne tareekon se A , B , C , D , E A B adjacent hon?
Glue karo A , B [ A B ] [ A B ] , C , D , E 4 ! = 24 4 kyun? Block ek count hota hai. Block ke andar A , B 2 ! = 2 Kyun? Adjacent hone par bhi 2 internal orders hote hain.
24 × 2 = 48
Worked example E4 — Apart (Pattern 3, gaps)A , B , C , D , E A B kabhi saath na hon .
Complement easy hai: Total = 5 ! = 120 = 48 120 − 48 = 72 Gaps check: C , D , E 3 ! = 6 4 _C_D_E_. A , B 4 P 2 = 12 6 × 12 = 72 4 gaps kyun? 3 objects se 3 + 1
Worked example E5 — Digits with restriction{ 1 , 2 , 3 , 4 , 5 } even hain?
Step 1 (constraint pehle): units digit even hona chahiye → { 2 , 4 } 2 Step 2: hundreds digit → 4 Step 3: tens digit → 3 2 × 4 × 3 = 24 Units pehle kyun? "Even" sirf units place restrict karta hai, isliye usse pehle lock karo warna freedom chhin jaayegi.
Common mistake "Together = sirf4 ! 2 ! Kyun sahi lagta hai: tumne correctly pair ko ek block mein glue kiya. Gap ye hai: block ke andar, do log phir bhi swap kar sakte hain. Fix: block ke internal arrangements ke liye r ! 2 !
n P r Kyun sahi lagta hai: n r n P r Trap: agar outcome ek set hai (ek committee, cards ki hand), to order irrelevant hai → combinations use karo n C r = r ! n P r Test: "Kya do chosen items ko swap karne se naya answer milega?" Agar nahi → combination.
Common mistake "Pehle free positions fix ki, phir constraint satisfy nahi ho payi"Kyun sahi lagta hai: left-to-right fill karna natural lagta hai. Fix: hamesha sabse restricted slot pehle handle karo (E5 units digit), warna zero valid choices reh sakti hain.
0 ! = 0 Kyun sahi lagta hai: "kuch ka factorial kuch nahi hona chahiye." Fix: 0 ! = 1 n P n = n ! n P 0 = 1
Recall Feynman: ek 12-saal ke bachhe ko samjhaoSocho tum apne doston ko photo ke liye line mein khada kar rahe ho. Pehli jagah ke liye tum unme se kisi ko bhi chun sakte ho. Ek baar koi wahan khada ho gaya, to agle ke liye ek kam rah jaata hai, aur aage bhi aisa hi. Tum multiply karte rehte ho "kitne bache hain." Agar koi left end par khade rehne ki zid kare , to unhe pehle wahan khada karo, phir baaki ko line karo — ye hai "restriction." Agar do best friends saath khade rehna chahein , to unhe ek invisible rope se baandh do aur ek insaan maano, lekin yaad rakho wo do taraf munh kar sakte hain (swap). Permutations = saari alag photo line-ups ginana!
Mnemonic Method yaad rakho"P is for Position, C is for Committee."
P ermutation → P laces/P ositions matter hain (order count hota hai).RESTRICT? → F irst the F orced, G lue if together, G aps if apart, C omplement for "at least/not". → "FF-GG-C"
n P r n r ordered arrangements ki sankhya, koi repetition nahi.
n P r n P r = ( n − r )! n !
n P n = n ! Kyunki n P n = ( n − n )! n ! = 0 ! n ! = n ! 0 ! = 1
n P 0 1
Permutation aur combination mein fark? Permutation order count karta hai (arrangements); combination order ignore karta hai (selections). n C r = r ! n P r
"Together" restriction ka method Objects ko ek block mein glue karo, blocks arrange karo, phir internal r !
"Apart" restriction ka method Gaps use karo: pehle baaki arrange karo, phir must-be-apart items ko unke beech ke gaps mein daalo (ya total − together use karo).
k k + 1
Restricted position ke liye kaunsa slot pehle fill karo? Sabse restricted position pehle, phir free positions fill karo.
{1,2,3,4,5} mein se even 3-digit numbers, no repeats 2 × 4 × 3 = 24
Fundamental Counting Principle — multiplication rule jis par sab kuch build hota hai.Combinations — nCr — same objects, order ignored; n C r = r ! n P r Factorials and 0! — 0 ! = 1 Circular Permutations — ring mein arrangements = ( n − 1 )! Permutations with Repetition — identical objects over-counts ko divide out karte hain.Probability — Equally Likely Outcomes — permutations aksar denominator dete hain.
Fill positions one by one
Permutation ordered arrangement
Restrictions change choices per step
Pattern 1 restricted position first
Pattern 2 together glue block
Pattern 3 apart gaps method
Pattern 4 fixed ends slots
Pattern 5 at least complement
