Question

Find the number of permutations of the letters of the word ASSASSINATION. In how many of these, 

(a) The vowels are in odd places. 

(b) Vowels are in odd places. 

(c) The word NATION is always present together. 

(d) Begins with ‘AS’ and ends with ‘AS’.

Answer 1

Total Letters = 13, A = 3, S = 4, I = 2, N = 2, T = 1, 0 = 1. 

This word can be arranged in = \(\frac{3!}{3!\times4!\times(2!)^2}\)

(a) The vowels are in even places 

The six vowels can be permuted in \(\frac{6!}{3!.2!}\)

There are 7 consonents, (S = 4, N = 2, T = 1), can be permutted in \(\frac{7!}{4!.2!}\)

The total number of ways = \(\frac{6!}{3!.2!}\). \(\frac{7!}{4!.2!}\)

(b) Vowels are in odd places: 1 2 3 4 5 6 7 8 9 10 11 12 13 14. 

The 7 consonents can be placed in 7 even places in \(\frac{7!}{4!.2!}\) ways and 7 odd places in which the six vowels have to be placed in ⁷P₆ ways for each of these 3 A’s and 21’s are repeated. 

∴ Vowels can be permuted in \(\frac{^6p_6}{3!.2!}\)

∴ Total number of ways = \(\frac{^6p_6}{3!.2!}\)  x \(\frac{7!}{4!.2!}\)

(c) The word NATION is always present together: Can be taken as 1 unit and remaining is 7 totally 8 can be done in \(\frac{8!}{2!.4!}\)

(d) Begins with ‘AS’ and ends with ‘AS’: The remaining 9 letters (SSINATION) can be permuted in \(\frac{9!}{2!.2!.2!}\)