How many words can be formed with the letters of the word 'FAILURE' in which consonants may occupy even positions?

Explanation

In the word 'FAILURE'
Consonants = F,L,R
Vowels = A,I,U,E
In the 7 positions
No. of even positions = 3
No. of odd positions = 4
No. of ways to fill 3 even positions with 3 consonants
`= .^(3)P_(3) = lfloor3 = 6`
No. of ways to fill 4 odd positions with 4 vowels
`= .^(4)P_(4) = lfloor4 = 24`
From multiplication theorem
Required permutations `= 6 xx 24 = 144`