A permutation is an ordered positions one by one , and at each position you count how many choices remain . Multiply those choices together — that's the whole subject. Restrictions just change how many choices you have at a given step .
An arrangement of objects in a definite order . The number of ways to arrange r r r n n n distinct objects (no repetition) is written = = n P r = = ==nP_r== == n P r ==
WHY order matters: A B AB A B B A BA B A different arrangements . If order did NOT matter (like picking a team), we'd be doing combinations , not permutations.
Fill r r r n n n
Position
Choices left
1st
n n n
2nd
n − 1 n-1 n − 1
3rd
n − 2 n-2 n − 2
… \dots … … \dots …
r r r n − ( r − 1 ) = n − r + 1 n-(r-1) = n-r+1 n − ( r − 1 ) = n − r + 1
By the multiplication principle , multiply the choices:
n P r = n ( n − 1 ) ( n − 2 ) ⋯ ( n − r + 1 ) nP_r = n(n-1)(n-2)\cdots(n-r+1) n P r = n ( n − 1 ) ( n − 2 ) ⋯ ( n − r + 1 )
Why this step? Each position is an independent decision given the earlier ones, and independent-and-then decisions multiply.
Now make it compact. Multiply and divide by the "leftover" factorial ( n − r ) ! (n-r)! ( n − r )!
n P r = n ( n − 1 ) ⋯ ( n − r + 1 ) ⋅ ( n − r ) ! ( n − r ) ! = n ! ( n − r ) ! nP_r = \frac{n(n-1)\cdots(n-r+1)\cdot(n-r)!}{(n-r)!} = \boxed{\dfrac{n!}{(n-r)!}} n P r = ( n − r )! n ( n − 1 ) ⋯ ( n − r + 1 ) ⋅ ( n − r )! = ( n − r )! n !
Almost every exam question is one of these five patterns . Master these and you've covered 80% of the marks.
WHY: the tightest constraint should be handled before you run out of freedom.
Treat the block as one super-object , arrange, then arrange inside the block.
Arrange the others first, then slot the "must-be-apart" ones into the gaps between them.
Place the forced objects, then permute the rest.
Count the total, subtract the forbidden.
Worked example E1 — Plainn P r nP_r n P r { A , B , C , D , E } \{A,B,C,D,E\} { A , B , C , D , E } 5 P 3 = 5 ! 2 ! = 5 ⋅ 4 ⋅ 3 = 60 5P_3=\frac{5!}{2!}=5\cdot4\cdot3=60 5 P 3 = 2 ! 5 ! = 5 ⋅ 4 ⋅ 3 = 60 Why 5 ⋅ 4 ⋅ 3 5\cdot4\cdot3 5 ⋅ 4 ⋅ 3 5 choices for slot 1, 4 for slot 2, 3 for slot 3.
Worked example E2 — Fixed position (Pattern 1)Arrange all of { A , B , C , D , E } \{A,B,C,D,E\} { A , B , C , D , E } starts with a vowel .
Step 1: first letter is a vowel → 2 2 2 A A A E E E Why first? It's the constraint. Step 2: arrange remaining 4 letters in 4 slots → 4 ! = 24 4! = 24 4 ! = 24 2 × 24 = 48 2\times 24 = 48 2 × 24 = 48
Worked example E3 — Together (Pattern 2, glue)In how many ways can A , B , C , D , E A,B,C,D,E A , B , C , D , E A A A B B B adjacent ?
Glue A , B A,B A , B [ A B ] [AB] [ A B ] [ A B ] , C , D , E [AB],C,D,E [ A B ] , C , D , E 4 ! = 24 4!=24 4 ! = 24 Why 4? The block counts as one. Inside the block A , B A,B A , B 2 ! = 2 2!=2 2 ! = 2 Why? Adjacent still has 2 internal orders.
24 × 2 = 48 24\times 2 = 48 24 × 2 = 48
Worked example E4 — Apart (Pattern 3, gaps)Arrange A , B , C , D , E A,B,C,D,E A , B , C , D , E A A A B B B never together .
Complement is easy: Total = 5 ! = 120 =5!=120 = 5 ! = 120 = 48 =48 = 48 120 − 48 = 72 120-48 = 72 120 − 48 = 72 Gaps check: arrange C , D , E C,D,E C , D , E 3 ! = 6 3!=6 3 ! = 6 4 4 4 _C_D_E_. Place A , B A,B A , B 4 P 2 = 12 4P_2 = 12 4 P 2 = 12 6 × 12 = 72 6\times12=72 6 × 12 = 72 Why 4 gaps? 3 objects create 3 + 1 3+1 3 + 1
Worked example E5 — Digits with restrictionHow many 3-digit numbers (no repeated digit) from { 1 , 2 , 3 , 4 , 5 } \{1,2,3,4,5\} { 1 , 2 , 3 , 4 , 5 } even ?
Step 1 (constraint first): units digit must be even → { 2 , 4 } \{2,4\} { 2 , 4 } 2 2 2 Step 2: hundreds digit → 4 4 4 Step 3: tens digit → 3 3 3 2 × 4 × 3 = 24 2\times4\times3 = 24 2 × 4 × 3 = 24 Why units first? "Even" only restricts the units place, so lock it before it steals your freedom.
Common mistake "Together = just4 ! 4! 4 ! 2 ! 2! 2 ! Why it feels right: you correctly glued the pair into one block. The gap: inside the block, the two people can still swap. Fix: multiply by r ! r! r ! 2 ! 2! 2 !
n P r nP_r n P r Why it feels right: you're choosing r r r n n n n P r nP_r n P r The trap: if the outcome is a set (a committee, a hand of cards), order is irrelevant → use combinations n C r = n P r r ! nC_r = \frac{nP_r}{r!} n C r = r ! n P r Test: "Would swapping two chosen items give a new answer?" If no → combination.
Common mistake "Fixed the free positions first, then couldn't satisfy the constraint"Why it feels right: left-to-right filling feels natural. The fix: always handle the most restricted slot first (E5 units digit), else you may leave zero valid choices.
0 ! = 0 0! = 0 0 ! = 0 Why it feels right: "factorial of nothing must be nothing." Fix: 0 ! = 1 0!=1 0 ! = 1 n P n = n ! nP_n=n! n P n = n ! n P 0 = 1 nP_0=1 n P 0 = 1
Recall Feynman: explain to a 12-year-oldImagine lining up your friends for a photo. For the first spot you can pick any of them. Once one stands there, you have one fewer to pick from for the next spot, and so on. You keep multiplying "how many are left." If someone insists on standing on the left end, put them there first , then line up the rest — that's a "restriction." If two best friends want to stand together , tie them with an invisible rope and treat them as one person, but remember they can face two ways (swap). Permutations = counting all the different photo line-ups!
Mnemonic Remember the method"P is for Position, C is for Committee."
P ermutation → P laces/P ositions matter (order counts).RESTRICT? → F irst the F orced, G lue if together, G aps if apart, C omplement for "at least/not". → "FF-GG-C"
What does n P r nP_r n P r The number of
ordered arrangements of
r r r objects chosen from
n n n distinct objects, no repetition.
Formula for n P r nP_r n P r n P r = n ! ( n − r ) ! nP_r = \dfrac{n!}{(n-r)!} n P r = ( n − r )! n ! .
Why is n P n = n ! nP_n = n! n P n = n ! Because
n P n = n ! ( n − n ) ! = n ! 0 ! = n ! nP_n=\frac{n!}{(n-n)!}=\frac{n!}{0!}=n! n P n = ( n − n )! n ! = 0 ! n ! = n ! since
0 ! = 1 0!=1 0 ! = 1 .
Value of n P 0 nP_0 n P 0 1 1 1 (one way to arrange nothing).
Difference between permutation and combination? Permutation counts order (arrangements); combination ignores order (selections).
n C r = n P r r ! nC_r=\frac{nP_r}{r!} n C r = r ! n P r .
"Together" restriction method Glue the objects into one block, arrange blocks, then multiply by internal
r ! r! r ! arrangements.
"Apart" restriction method Use gaps: arrange the others first, then place the must-be-apart items into the gaps between them (or use total − together).
Number of gaps from k k k k + 1 k+1 k + 1 gaps (including the two ends).
For a restricted position, which slot do you fill first? The most restricted position first, then fill the free positions.
Even 3-digit numbers from {1,2,3,4,5} no repeats 2 × 4 × 3 = 24 2\times4\times3=24 2 × 4 × 3 = 24 .
Fundamental Counting Principle — the multiplication rule everything is built on.Combinations — nCr — same objects, order ignored; n C r = n P r r ! nC_r=\frac{nP_r}{r!} n C r = r ! n P r Factorials and 0! — why 0 ! = 1 0!=1 0 ! = 1 Circular Permutations — arrangements in a ring = ( n − 1 ) ! =(n-1)! = ( n − 1 )! Permutations with Repetition — identical objects divide out over-counts.Probability — Equally Likely Outcomes — permutations often give the denominator.
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
Intuition Hinglish mein samjhoDekho, permutation ka matlab hai order-wale arrangement . Jab tum cheezon ko ek line me rakhte ho, to har position ko bhar-te jao aur socho "ab kitne choices bache hain". Pehli jagah ke liye n n n n − 1 n-1 n − 1 n − 2 n-2 n − 2 n P r = n ! / ( n − r ) ! nP_r = n!/(n-r)! n P r = n ! / ( n − r )! ek option kam ho jaata hai kyunki repeat allowed nahi.
Ab restrictions — yahi exam me marks laate hain. Agar koi position fix hai (jaise "vowel se start"), to wahi constraint pehle solve karo , baaki ko baad me arrange karo. Agar do cheezein saath rehni hain, unko ek rassi se baandh do (glue), ek object maan lo, phir andar ka 2 ! 2! 2 ! alag rehni hain, to baaki ko pehle arrange karo aur gaps me daal do — ya seedha "total minus saath-wale" nikaal lo.
Sabse badi galti: order matter karta hai ya nahi, ye check karna. Committee/team choose karni ho to order nahi chalta — wahan combination (n C r nC_r n C r 0 ! = 1 0!=1 0 ! = 1 n P n = n ! nP_n=n! n P n = n ! n P 0 = 1 nP_0=1 n P 0 = 1
