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Permutations of distinct objects

A permutation is an arrangement of objects in a definite order. If you change the order, you get a different permutation. So ABCDABCD and BACDBACD are different permutations of the four letters A,B,C,DA, B, C, D.

Definitions

The number of permutations of nn distinct objects taken rr at a time is denoted P(n,r)P(n, r) (or nPr^nP_r, or PrnP^n_r). It is the number of distinct sequences of length rr whose entries come from a set of nn distinct items, with no repetition.

By the multiplication principle: P(n,r)=n(n1)(n2)(nr+1)=n!(nr)!.\boxed{P(n, r) = n \cdot (n - 1) \cdot (n - 2) \cdots (n - r + 1) = \dfrac{n!}{(n - r)!}.}

Special cases:

  • P(n,0)=1P(n, 0) = 1 (the empty sequence).
  • P(n,n)=n!P(n, n) = n! (full arrangement of all nn).
  • P(n,1)=nP(n, 1) = n.

Derivation

To form a sequence of length rr from nn distinct items:

  • 1st1^{\text{st}} position: nn choices.
  • 2nd2^{\text{nd}}: n1n - 1 choices (one already used).
  • \vdots
  • rthr^{\text{th}}: nr+1n - r + 1 choices.

Total: n(n1)(nr+1)=n!(nr)!n (n - 1) \cdots (n - r + 1) = \dfrac{n!}{(n - r)!}. \qed\qed

Common applications

  • Words / letters / codes. Arrange rr letters from nn.
  • Numbers. Form rr-digit numbers from nn distinct digits.
  • Queues / seats. Arrange rr people in rr specific seats from a pool of nn.

Permutations under constraint

A typical Class-XI problem adds a constraint:

  • Two specific objects together. Treat them as a block (Example 4 of subtopic 1).
  • Two specific objects apart. Subtract "together" cases from total.
  • Specific position. Fix the constrained position first, then arrange the rest.

Worked examples

Example 1. How many 33-letter sequences can be formed from A,B,C,D,E,F,GA, B, C, D, E, F, G if (a) repetition is allowed, (b) not allowed?

(a) Multiplication principle: 73=3437^3 = 343.

(b) P(7,3)=765=210P(7, 3) = 7 \cdot 6 \cdot 5 = 210.

Example 2. In how many ways can the letters of the word "PEACE" be arranged in a row? (We will revisit identical letters in the next subtopic; for now treat each letter as distinct, i.e. just count arrangements of 55 distinct symbols.)

If treated distinct: 5!=1205! = 120. (If EE's are identical, 5!2!=60\dfrac{5!}{2!} = 60.)

Example 3. Find nn such that P(n,5)=20P(n,3)P(n, 5) = 20 \cdot P(n, 3).

n!(n5)!=20n!(n3)!\dfrac{n!}{(n - 5)!} = 20 \cdot \dfrac{n!}{(n - 3)!}.

Divide both sides by n!(n3)!\dfrac{n!}{(n - 3)!} (non-zero): (n3)!(n5)!=20    (n3)(n4)=20.\frac{(n - 3)!}{(n - 5)!} = 20 \iff (n - 3)(n - 4) = 20. So n27n+12=20n^2 - 7n + 12 = 20, i.e. n27n8=0n^2 - 7n - 8 = 0, n=8n = 8 (rejecting n=1n = -1).

Example 4. How many 44-digit numbers without repeated digits can be formed from 1,2,3,4,5,6,71, 2, 3, 4, 5, 6, 7?

P(7,4)=7654=840P(7, 4) = 7 \cdot 6 \cdot 5 \cdot 4 = 840.

Example 5 (harder). How many arrangements of the letters of "BANANA" (with repeats) have the two NN's never together?

The word has letters B,A,N,A,N,AB, A, N, A, N, A , six letters, but three AA's and two NN's (and one BB). Total distinct arrangements: 6!3!2!=60\dfrac{6!}{3! \cdot 2!} = 60.

Arrangements where the two NN's are together: glue them as a block "NNNN". We are arranging 55 objects: B,A,A,A,NNB, A, A, A, NN. That gives 5!3!=20\dfrac{5!}{3!} = 20.

Hence number with NN's not together: 6020=4060 - 20 = 40.

Try it yourself

  1. Compute P(10,3)P(10, 3), P(7,7)P(7, 7), P(6,4)P(6, 4).
  2. How many 55-digit numbers can be formed from 1,2,3,4,5,6,7,81, 2, 3, 4, 5, 6, 7, 8 with no repetition?
  3. Find nn: P(n,4)=12P(n,2)P(n, 4) = 12 \cdot P(n, 2).
  4. In how many ways can 66 people stand in a row?
  5. In how many of these does person AA stand first?
  6. In how many ways can the letters of "EQUATION" be arranged so that all vowels are together?
  7. How many 44-letter words start with a consonant from the letters A,B,C,D,EA, B, C, D, E?
  8. How many 55-digit numbers from 0099 without repetition have no leading zero?
  9. How many permutations of "MATHEMATICS" treating letters as distinct? (Then divide by repetitions: subtopic 4.)
  10. In a class of 2525, in how many ways can a President, Vice-President and Secretary be chosen?
  11. How many ways can 44 red and 33 blue balls be arranged in a row if all balls are distinguishable?
  12. How many three-digit even numbers with no repeated digits can be formed from 1,2,3,4,5,61, 2, 3, 4, 5, 6?

Pitfalls / Tricks

  • Permutations are about order. ABCACBABC \ne ACB as permutations even though the same letters are used.
  • Always check: are objects distinct? Is repetition allowed? Two questions before any formula.
  • Constraint-based problems: handle the constraint first, then arrange the rest freely.
  • Insight. When two objects must stay together, glue them and multiply by the internal orderings. When two must stay apart, count "all" minus "together".

Practice quiz

Quick check on this topic.

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Quick check : Permutations of distinct
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