Aptitude - Permutation and Combination - Discussion

In how many ways can the letters of the word CIVIL be rearranged? Can anyone explain?

Please.

In how many ways can the letter of word failure be arranged si that the consonants may occupy only odd position.

How there are three vowels in the word vowels? logically we cannot consider the word "vowels" as a vowel. Can anyone explain, please?

In how many ways can the letters of the word SYSTEM be arranged so that E is always at the third position?

Why we also considering the positions of the consonants?

It is 36. Assuming we represent the filings as D*T**L.
So that d (*) the possible fillings for the vowels (E A I).
Then "E" can go into the arrangement in 3ways I. E either "E" takes d first, second, or the third filling space ie shown by the (*) sign.
So after E as occupying one space, then we are left with just spaces left which are to be filled by "A" and "I". In which "A" can go into it in 2ways. Remaining one space to be occupied by "I" in just 1way.

So arrangement of vowel is 3P3 = 3! = 3 * 2 * 1 = 6. So we have 3 consonant (D T L) left we can permute (arrange) them in 3! ways = 3 * 2 * 1 = 6 ways. So total arrangement = 6 * 6 = 36 ways.

There are 6 positions out of which 1, 3 and 5 are for the three vowels. There are 3 ways for arranging in position 5 and remaining 2 ways for position 3 and 1 way for position 1. So for vowels, it is 3 * 2 * 1 = 6.
For remaining three consonants which are placed in even position, there are 3 ways in position 6 and remaining 2 ways in position 4 and one way in position 2. So for consonants, it is 3 * 2 * 1 = 6.

So a total number of ways = 6 * 6 = 36.

@Devdp.

I too agree with you. If I name the places from left-hand side i.e. 6-5-4-3-2-1 or from right-hand side 1-2-3-4-5-6. I will get 2 types of arrangement and since the question is regarding permutation both way of arrangement matters, so answer we get is 72.

@Aditya.

I have the solution for your question you have the word "FATHER" that is nothing but 6 letters in which 2 vowels and 4 consonants. So we have to arrange the vowels at odd places as we know that out of 6 words 3 places are odd places, so the arrangement of 2 vowels at 3 places means 3C2 right.

So we arranged that vowel at odd places now we have to take the permutation of that vowels as 2! and the permutation of the consonants that is nothing but 4! so finally we have the solution as :-
=> 3C2 * 2! * 4!
=> 3 * 2 * 24.
=> 144.

In how many ways we can arrange the alphabets of the word 'ARRANGE' such that 2a and 2r do not come together?

Please solve this.