# Solved Examples on Permutations

Illustration 1: Match the conditions in column 1 with the most suitable descriptions in column 2. (2008)

Consider all possible permutations of the letters of the word ENDEANOEL:

 Column 1 Column 2 1. The number of permutations containing the word ENDEA, is 1. 5! 2. The number of permutations in which the letter E occurs in the first and last positions is, 2. 2x5! 3. The number of permutations in which none of the letters D, L, N occurs in the last five positions is, 3. 7x5! 4. The number of permutations in which the letters A, E, O occur only in odd positions is, 4. 21x5!

Solution: We consider all the parts one by one.

(1) If ENDEA is a fixed word, then we assume it as a single letter. Hence, the total number of letters is 5. Hence, the total number of arrangements is = 5!

(2) Now, if the letter ‘E’ is at first and last places, then the required number of permutations are

7!/2! = 21.5!

(3) Now, the letters D, L and N are not in last five positions

So, ← D, L , N, N → ← E, E, E A, O →

Hence, total number of permutations = 4!/2! X 5!/3! = 2.5!

(4) Total number of odd positions = 5

Permutations of AEEEO are 5!/3!

Total number of even positions = 4

Number of permutations of N, N, D, L = 4!/2!

Hence, total number of permutations = 5!/3! . 4!/2! = 2.5!

Illustration 2: Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit on one particular side and the rest three on the other side. Determine the number of ways in which the seating arrangement can be done. (1991)

Solution: Let the two sides be A and B. Now, let us assume that the four particular guests wish to sit on side A. So, these 4 guests can be accommodated on nine chairs in 9P4 ways. Similarly, the three guests who wish to sit on side B can be accommodated in 9P3 ways.

So, now 11 guests are left who can sit on remaining chairs in 11! ways. Hence, the total number of ways in which 18 persons can be seated on chairs is given by

9P4 .9P3 .11!

Illustration 3: Two planes P1 and P2 pass through origin. Two lines L1 and L2 also passing through origin are such that L1 lies on P1 but not on P2, L2 lies on P2, but not on P1. If A, B, C are three points other than origin then prove that the permutation [A’B’C’] of [ABC] exists such that

(a) A lies on L1, B lies on P1 not on L1, C does not lie on P1.

(b) A’ lies on L2, B’ lies on P2 not on L2, C’ does not lie on P2. (2004)

Solution: In this case, a total of 6 permutations are possible. A corresponds to one of A’, B’, C’ and B corresponds to one of remaining A’, B’, C’ and C corresponds to the third of A’, B’, C’.

This implies that A lies on L1, B lies on the line of intersection of P1 and P2 and C lies on the line L2 on the plane P2.

Therefore, A’ lies on L2 = C

B’ lies on the line of intersection of P1 and P2 = B

C’ lies on L1 on the plane P1 = A.

Thus, we get [A’B’C’] = [CBA]

Hence, Permutation of [A’B’C’] is [ABC].

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