Aptitude - Permutation and Combination - Discussion

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"In the middle of difficulty lies opportunity."

- Albert Einstein

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

Answer: Option A

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters = 7! = 2520.

2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in 5! = 20 ways.

3!

Required number of ways = (2520 x 20) = 50400.

Pritesh Patel said: (Thu, Sep 23, 2010 04:49:15 AM)

Why we dividing 7! and 5! by 2! and 3!, respectively. Is there any formula for that? What is the Logic Behind it? Thanks.

Satty said: (Tue, Oct 12, 2010 05:06:29 PM)

I have the same question, Is there a formula behind ?

Priyanka said: (Sat, Oct 16, 2010 12:04:53 AM)

I also can't understand division. Is there any formula ?

Nishtha Sharma said: (Wed, Dec 1, 2010 11:20:27 PM)

Here we can simply separate out the common alphabet. in CORPORATION , o is 3 times and r is two times, putting together we have,OOORRCPATIN . now the total unique alphabets are 7 and thus the answer is 7%=7654321= 50400 . the common alphabets dun need any permutation.

Venumadhav said: (Tue, Dec 28, 2010 11:29:06 AM)

Actually if you consider ooo as o and rr as are the there are 8 distinct alphabets. C O are P A T I N.

Kumaran said: (Thu, Jan 6, 2011 02:41:08 AM)

How the 7 is came instead of 6 (CRPRTN).

Raghav said: (Wed, Jan 26, 2011 07:50:46 AM)

What is process when same letter is come in like as this question? Thanks.

Suja said: (Sat, Feb 12, 2011 08:37:17 AM)

We consider all vowels as one letter i.e. OOAIO is a letter + rest of the 6 letters = 1+6 =7.

Vinod said: (Sat, Jul 23, 2011 10:27:51 PM)

Why divide 7! by 2! ? because you can treat the problem as a permutation with subgroups of identical items. the general formula is nPn1,n2,n3... equals n! divided by n1!n2!n3!... In this problem you have n = 7 letters (6 plus the vowel group). two letters are the same so n1 = 2. the rest are unique so the 5 other subgroups = 1. so you 7! divided by 2!1!1!1!1!1! . the answer as given simply didn't write out the 1! terms.

Kanakam Vinay Kumar said: (Thu, Aug 11, 2011 11:15:44 AM)

I understood Very Clearly.