Combination Question?

Question

Did I solve this combination question correctly? I want to know if I finally understand how to do combinations. And if it's wrong, can someone please explain it? The problem is...

Sets of 4 letters are chosen from the English alphaebet. Find the number of 4-letter sets possible if there must be the same number of vowels and consonants, and if A is always included.

The setup I got for this problem was 4C1 * 21C2. And my answer is 840.

Answer 1

we will solve using y as a consanant first, then with y as a vowel.

y= consanant there are 5 vowels and 21 cosanants

same number of vowels and consanants means there must be 2 of each

as a combination (A,B,C,E = E,C,B,A = A,E,C,B )

there are 4C1*21C2 = 840 combinations

( always have an A so pick 1 of 4 vowels and 2 of 21 consanants order does NOT matter )

if y is a vowel then it is

(5C1)(20C2) = 950 combinations

Answer 2

Pi is right.

The man with (coincidentally) no vowels in his name is wrong.

His answer would only be correct if the order mattered. Since the order of the consonants doesn't matter, there are only half that many.

Answer 3

There must be 2 vowels and 2 consonants. One of the vowels must be A so there are 4 possibilities for the second vowel. Then there are 21 possibilities for the first consonant since there are 26-5 = 21 consonants. Then once that consonant is chosen ,there are 20 possibilities for the second consonant. So the answer is:

4 * 21 * 20 = 1680.

Answer 4

Hi,

If A is always included as one of the 4 letters, then there are really 3 other letters to fill in. One is a vowel so it can be any of the other 4 vowels. The other 2 spots must be consonants, so they are any 2 of 21 consonants.  These spots could be filled in this many ways:

4*(21nCr2) = 4* 210 = 840 <==ANSWER

Your answer is correct!! Good job!!

I hope that helps!! :-)