Aptitude - Permutation and Combination - Discussion

SIMPLE TRICK:

There are 6 Letters 3 V and 3 C.

6blank positions_ _ _ _ _ _.

Which can be filled like this VCVCVC.

3V3C2V2C1V1C = 3 * 3 * 2 * 2 * 1 * 1 = 36.

@Keshav.

By Using the Gauss method. The correct answer is A.

@Devdp.

As per the question, Vowels can occupy only the odd position. But you have placed the vowel in even position. Explain clearly.

DETAIL = vowels ( E A I) = 3!
No.of possible is(DETAIL)
_ _ _ _ _ _
1 2 3 4 5 6 letter so conditions is odd position only 1,3,5,place only possible....=3!
Then 3! x 3! = 36.

Here question says vowels occupy only the odd position, then what about consonants? Please clarify this.

Simple:

(DTL) * (EAI).

(3*2*1) * (3*2*1).
(6)*(6) = 36.

We can arrange consonants in 3! ways in even position and 4 odd places can be occupied by 3 vowels in 4P3 ways.

So these can be arranged in 3!*4P3 =144 ways.

Am I right?

Please give me the answer.

What if the question is,

In how many different ways can the letters of the word "PUNCTUATE" be arranged in such a way that the vowels occupy only the even positions? Observe the vowels: UUAE (here 'U' occurs twice).

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.

@Shekhar.

Civil = 5!/2! = 60.