Aptitude - Permutation and Combination - Discussion

Why making this question so complicated?

See we have the word "DETAIL" which is having the total number of alphabets that is "6".

In which "3" is vowels and "3" is consonants and we have to arrange the vowels at odd places right so what are the positions we have (1) 2 (3) 4 (5) 6 [number inside the braces indicates the odd places].

So we have 3 places and we need to arrange the 3 vowels we do it by combination that is 3c3=1 and the no of vowels is 3 so their permutation is 3! so that the similar permutation of consonants is having the same 3! because we have 3 consonants as well so finally the answer is

=> 3C3 * 3! * 3!

=> 1* 6 * 6.

=> 36.

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.

@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.

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}.

Total number of ways = (6 x 6) = 36.

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.

The word detail can be arranged with vowels at odd places in 3C1*3C1*2CI*2C1*1C1*1C1 as no of vowels are 3 and consonants are 3 in no.

So 1st place can be arranged in 3C1 ways. Similarly, 2nd place can be arranged in 3C1 ways as there are 3 consonants. Similarly, the 3rd 4th 5th and 6th place can be arranged in 2C1, 2C1, 1C1, 1C1 ways respectively. Total no of ways is their product. This problem can be done with combination also. Am I right?

When 1st try myself consider x vowel and y consonants it can be possibility of position (1) xyxyxy and (2) yxyxyx.

As per permutation for x 3p3 = 3! = 6.

Permutation for why 3y3 = 3! = 6.

And as shown as above 2 different position.

Total arrangement = 6*6*2 = 72 (Ans).

But when I shows option I confused. There is no 72.

If we can set the vowels in odd positions, we can get one combination like;

ED AT IL.

Like that;
So always one consonant comes along with a vowel.
So total combined words= 3!
And 3 vowels can arrange themselves in 3! -> Number Of ways.
So, total arrangement is= 3! *3!.
=36.

Since there is nothing mentioned about repetitions so the no ways could also be 3*3*3(i.e., E can come in all three odd places and same goes for the other two vowels A and I as well) for the three odd positions of vowels.

And similarly the other three letters can be arranged so the answer should be 3*3*3*3*3*3.

@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.