Aptitude - Permutation and Combination - Discussion

This question is only based on a combination, they never told to find the arrangements. So the answer is 210.

Why can't we use permutation here instead of combination and multiplication by arrangement?

Since permutation = combination * arrangement.

But the answer is coming 2520.

As the consonants and vowels are not specified and we only have to find how many ways we can arrange them i.e. 210 the Order does not matter. If the letters were specified then only we can apply permutation here.

For example, if the question was 7 men and 4 women, we wouldn't multiply with 5!.
Hence, the correct answer would be 210.

@Veyron999.

Best explanation.

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.

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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?

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