Introduction to Permutations

A permutation is an arrangement of a number of objects in a definite order. In permutations, the order of the objects is crucial.

Factorial Notation

Factorial notation is essential for calculating permutations. For a natural number n, the factorial of n, denoted by n!n!, is the product of the first n natural numbers.

n!=n×(n−1)×(n−2)×...×2×1n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1

  • By definition, 0!=10! = 1.

1. Permutations of Distinct Objects (Without Repetition)

The number of permutations (arrangements) of n distinct objects taken r at a time is denoted by nPr^nP_r or P(n,r)P(n,r).

nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}

  • The number of permutations of n distinct objects taken all at a time is nPn=n!^nP_n = n!.

2. Permutations with Repetition

The number of permutations of n distinct objects taken r at a time, when repetition is allowed, is nrn^r.

3. Permutations of Non-Distinct Objects

The number of permutations of n objects, where there are p1p_1 objects of one kind, p2p_2 objects of a second kind, …, and pkp_k objects of a k-th kind, is given by:

n!p1!⋅p2!⋅...⋅pk!\frac{n!}{p_1! \cdot p_2! \cdot ... \cdot p_k!}

4. Circular Permutations

The number of ways to arrange n distinct objects around a circular table is (n−1)!(n-1)!. This is because in a circle, there is no fixed starting point, so we fix the position of one object and arrange the remaining (n-1) objects.

Example 1: Using the nPr^nP_r Formula

Question: How many 4-letter words, with or without meaning, can be formed out of the letters of the word 'LOGARITHMS', if repetition is not allowed?

Solution: The word 'LOGARITHMS' has 10 distinct letters. We need to arrange 4 of them. This is a permutation of 10 objects taken 4 at a time. The number of ways is 10P4=10!(10−4)!=10!6!=10×9×8×7=5040^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040.

Example 2: Permutations with a Constraint (Vowels Together)

Question: Find the number of different 8-letter arrangements that can be made from the letters of the word 'DAUGHTER' so that all vowels occur together.

Solution: The word 'DAUGHTER' has 8 distinct letters, including 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R).

  1. Treat vowels as one unit: Consider the block of vowels (AUE) as a single object. Now we have 5 consonants + 1 block of vowels = 6 objects to arrange. These can be arranged in 6!6! ways.
  2. Arrange vowels internally: The 3 vowels within their block can be arranged in 3!3! ways. By the multiplication principle, the total number of arrangements is 6!×3!=720×6=43206! \times 3! = 720 \times 6 = 4320.

Example 3: Permutations of Non-Distinct Objects

Question: In how many ways can the letters of the word 'MISSISSIPPI' be arranged?

Solution: The word has 11 letters in total. The letter M appears 1 time, I appears 4 times, S appears 4 times, and P appears 2 times. Using the formula for non-distinct objects, the number of arrangements is:

11!1!⋅4!⋅4!⋅2!=399168001⋅24⋅24⋅2=34650\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39916800}{1 \cdot 24 \cdot 24 \cdot 2} = 34650

Example 4: Circular Permutations

Question: In how many ways can 7 people be seated around a circular table?

Solution: For a circular arrangement of n distinct objects, the number of ways is (n−1)!(n-1)!. Here, n=7. The number of ways is (7−1)!=6!=720(7-1)! = 6! = 720.