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 \(\frac{9!}{2!.2!.2!}\)