Aptitude - Permutation and Combination - Discussion

You have to arrange the letter.

3 consonants out of 7.

2 vowels out of 4.

7c3x4c2 = 210.

You can do,

How many letter out of these letters?

3+2 = 5.

Now the required no of ways = 210 x 5!

210 is correct. What is asked in the problem is how many words that have 3 consonants and 2 vowels can be formed from the given 7 and 4.

Basically all words that contains 3 consonants no matter how you arranged the 3 consonants that you picked is considered as one, same with the vowels. Meaning QEWAR is same with WEQAR, hence counted as 1 sample.

Unless you choose another consonant say QEWAD or another vowel QIWAR then it will be a different sample.

I don't understand it.

In question instead of how many words there should be how many arrangements.

Otherwise, a solution is wrong according to question.

Explanation is actually simple. Suppose if they asked to find the number of words possible using five different letters and you had no constraints of consonants and vowels.

Then the answer would obviously 5!. So, we know you can form 5! words from five letters.

Now, in the problem, there are not just 5 letters, but much more - 7 consonants and 4 vowels. The 3 consonants can be selected in 7C3 ways. For each of these, we can select 2 vowels in 4C2 ways.

So, the 5 letters themselves can be selected in 7C3X4C2 ways.

Note that each of these 7C3X4C2 are a different set of five letters. A given set of 5 letters can be arranged in 5! ways.

So, 7C3X4C2 sets of 5 letters can be arranged in 7C3X4C2X5! ways.

Hence the answer is 25,200