# Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 2)

2.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

Answer: Option

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Video Explanation: https://youtu.be/WCEF3iW3H2c

Discussion:

97 comments Page 1 of 10.
Shoba said:
1 decade ago

Other than vowels there are only 4 letter then how it s possible to get 5!.

Sai said:
1 decade ago

HI SHOBA,.

4 consonants + set of vowels (i. E. , L+N+D+G+ (EAI) ).

We should arrange all these 5. So we get 5!.

I think you understood.

4 consonants + set of vowels (i. E. , L+N+D+G+ (EAI) ).

We should arrange all these 5. So we get 5!.

I think you understood.

Sachin Kumar said:
1 decade ago

Please make me understand this answer as did not get.

Sri said:
1 decade ago

As vowels are together take (EAI) as single letter i.e. , total no of letters are 5 (L, N, D, G, {EAI}).

No of ways can arrange these 5 letters are 5! ways.

Now we arranged 5 letters (L, N, D, G, {EAI}).

Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).

All these combinations imply that vowels are together.

So we have to multiply 5! and 3!.

No of ways can arrange these 5 letters are 5! ways.

Now we arranged 5 letters (L, N, D, G, {EAI}).

Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).

All these combinations imply that vowels are together.

So we have to multiply 5! and 3!.

Subbu said:
1 decade ago

7!=5040

Madhusudan said:
1 decade ago

Could you kindly let me know what is ( ! ).Howe 5! = 120 ?

Sundar said:
1 decade ago

@Madhusudan

5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120

5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120

Laxmikanth said:
1 decade ago

When should we take the one or more letters as a single unit and why?

Moaned said:
1 decade ago

I am not getting

Jessie said:
1 decade ago

7 letter word = LEADING

CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD

SO NO. OF WORDS = L,(EAI),D,N,G = 5

permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!

0! is assumed to be 1!)

EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!

hence 5! x 3! = 120 x 6 = 720

CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD

SO NO. OF WORDS = L,(EAI),D,N,G = 5

permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!

0! is assumed to be 1!)

EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!

hence 5! x 3! = 120 x 6 = 720

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