Question

How many words that can be formed with the letters of the word ‘SWIMMING’ such that the vowels do not come together? Assume that words are of with or without meaning.

(a) 430

(b) 623

(c) 729

(d) 1239

I had been asked this question in a job interview.

Origin of the question is Counting topic in division Counting of Discrete Mathematics

Answer 1

Correct answer is (c) 729

Easiest explanation: The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice. Therefore, the number of words formed by this word = \(\frac{8!}{2!*2!}\) = 10080. In order to find the number of permutations that can be formed where the two vowels I and I come together,  we group the letters that should come together and consider that group as one letter. So, the letters are S, W, M, M, N, G, (I, I). So, the number of letters are 7 the number of ways in which 7 letters can be arranged is 7! = 5040. In I and I, the number of ways in which I and I can be arranged is 2!. Hence, the total number of ways in which the letters of the ‘SWIMMING’ can be arranged such that vowels are always together are \(\frac{7!}{2!*2!}\) = 5040 ways. The number of words in which the vowels do not come together is = (10080 – 5040) = 5040.