Aptitude - Permutation and Combination - Discussion

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?

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!.

The correct answer should be option A.

I agree @Ankita.

That as we are picking up the words by applying combination means that we are forming the words therefore answer. It Should be 210.

@Himatheja.

We have to make word from both consonants and vowel that's why we have to multiply combination of consonants with vowel also,
7c3*4c2.
= 35*6.
= 210.

4C2 = (3*4)/(2! * 2!).
2 VOWELS OUT OF 4.
then, why 210?

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?

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

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.

It's.

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