Aptitude - Permutation and Combination - Discussion

3 out of 7 consonants,
7 * 6 * 5/3 * 2 * 1 = 210/6
2 out of 4 vowels 4 * 3/2 * 1 = 12/2 = 6,
210/6 * 6 = 210.

Formation of 3 Consonants and 2 vowels 3 + 2 = 5.
5 * 4 * 3 * 2 * 1 = 120,
210 * 120 = 25200.

It's.

VCVCC
VCVCV
VCCVC
VCCCV
CVVCC
CCVVC
CCVVC
CCVCV
CCcvv

It's simple,

Select 3 consonants out of B,C,D,F,G,H,J (assume these 7 consonants) in 7P3 ways.
So, [ B C D ] is one selection out of 7P3 ways right?
but, [ C B D] , [ C D B ] , [ B D C ] , [ D C B ] , [ D B C ] , all these contain same consonants.

This is why you use the combination, to form groups of 3 consonants out of 7 consonants.
Same goes for vowels : 4P2 ways to get groups of 2 vowels out of 4 vowels.
total groups of consonants and vowels : 7P3 * 4P2.
Now, we need to form words of 5 letters consisting of vowels and consonants.
How many ways? In 5! ways.

Thus, 7P3 * 4P2 * 5! = 25200 ways.

I don't think that it's supposed to be multiplied by 5! since it doesn't ask for the arrangement of the letters.

How many numbers greater than 5000 can be formed with the digits, 3, 5, 7, 8, 9 no digit being repeated?

In this question we won't multiply instead of adding?

How can we find permutation and combination problem ?

Hi,

I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.

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

Here, the answer is 302400.

Because you should 3 constants from 7 and 2 vowels from 4 and multiply then the answer is (7*6*5) * (4*3).


Later you should arrange them by multiplying with 5!.

I'm not sure this is the correct forum for this question, but here goes. If I have 4 consonants and there are 5 vowel sounds that can be pronounced hard or soft, how many different pronunciations are possible?

If this is the wrong forum can you direct me to a place that could answer that question?