Let W6 letter words with 2 or 4 vowels not including y All t

Let W={6 letter words with 2 or 4 vowels not including y}

All the letters are in lower case and repeates are okay.

What is the size of W?

Solution

SOLUTION

Back-up Theory

1. Any r out of n things can selected in ⁿC_r ways if repetition is not permitted and in n^r ways if repetition is permitted and ⁿC_r = (n!)/{(r!)(n - r)!}

2. r things can be arranged (or permuted) in (r!) ways if repetition is not permitted and in r^r ways if repetition is permitted.

3. Word formation is always a permutation problem.

4. In given problem, since y is excluded, total number of letters available is only 25.

Working

Case 1: 6-letter word has exactly 2 vowels.

=> the word has 2 vowels and 4 consonants => selection has to be 2 out of 5 vowels and 4 consonants out of 20 consonants (note: y cannot be included). Since repetition is allowed , number of possible selections = 5²20⁴ and each of these selections are to permuted to get the desired number of words. This has to be done for various possibilities since repetition is allowed.

(1): vowels and consonants are distinct => number of permutations = 6!

(2): vowels are same and consonants are distinct => number of permutations = (6!)/(2!)

(3): vowels are distinct and out of 4 consonants 2 are same 2 are distinct distinct => number of permutations = (6!)(⁴C₂)/(2!)

(4): vowels are distinct and out of 4 consonants 3 are same 1 is distinct distinct => number of permutations = (6!)(⁴C₃)/(3!)

(5): vowels are distinct and out of 4 consonants all are same => number of permutations = (6!)/(4!)

(6): vowels are same and out of 4 consonants 2 are same 2 are distinct distinct => number of permutations = (6!)(⁴C₂)/{(2!)(2!)}

(7): vowels are same and out of 4 consonants 3 are same 1 is distinct distinct => number of permutations = (6!)(⁴C₃)/ {(2!)(3!)}

(8): vowels are same and out of 4 consonants all are same => number of permutations = (6!)/{(2!)(4!)}

The number of possible words under Case 1 = 5²20⁴ x A,

where A = sum of all results in (1) to (8) = (6!){1 + (1/2) + 3 + (2/3) + (1/24) + (3/2) + (1/3) + (1/48)} = (6!)(113/16) = 45.

Thus, Case 1 total = 18x10⁷

Case 2: 6-letter word has exactly 4 vowels.

In this case, number of selections = 5⁴20²

Here again, various possibilities as above can be worked in identical manner, but it is not necessary since vowels are replaced by consonants and vice-versa. Hence A = 45 in this case also

Thus, Case 2 total = 1125x10⁴

AND FINAL ANSWER = SUM OF CASE 1 AND CASE 2.