Problem on permutation & combination

Pinku wrote:The number of words that can be formed out of the letters a,b,c,d,e,f taken 3 together each word containing atleast one vowel.
a)96 b)90 c)24 d)120 e)12

In future, please don't paraphrase questions - make sure you reproduce the exact wording. In this case we can understand what the question means, but sometimes inaccurate reproduction makes questions unanswerable. Further, please always provide the source.

On to the question!

Each word must contain at least one vowel, so there are two possible cases:

1 vowel + 2 consonants; or
2 vowels + 1 consonant.

Case 1:

2C1 * 4C2 = 2 * (4*3/2) = 2 * 6 = 12 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 12*6 = 72 possible words.

Case 2:

2C2 * 4C1 = 1 * 4 = 4 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 4*6 = 24 possible words.

Either case is acceptable, so we ADD them together:

72 + 24 = 96... choose (A).

Stuart, have a doubt here.

Can't we chose the same vowel twice. Like 'eed' or 'aaf'
or have 'aaa', 'eee', 'aae' etc ?
Shouldn't this be considered too?

Stuart Kovinsky wrote:

Pinku wrote:The number of words that can be formed out of the letters a,b,c,d,e,f taken 3 together each word containing atleast one vowel.
a)96 b)90 c)24 d)120 e)12

In future, please don't paraphrase questions - make sure you reproduce the exact wording. In this case we can understand what the question means, but sometimes inaccurate reproduction makes questions unanswerable. Further, please always provide the source.

On to the question!

Each word must contain at least one vowel, so there are two possible cases:

1 vowel + 2 consonants; or
2 vowels + 1 consonant.

Case 1:

2C1 * 4C2 = 2 * (4*3/2) = 2 * 6 = 12 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 12*6 = 72 possible words.

Case 2:

2C2 * 4C1 = 1 * 4 = 4 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 4*6 = 24 possible words.

Either case is acceptable, so we ADD them together:

72 + 24 = 96... choose (A).

nithi_mystics wrote:Stuart, have a doubt here.

Can't we chose the same vowel twice. Like 'eed' or 'aaf'
or have 'aaa', 'eee', 'aae' etc ?
Shouldn't this be considered too?

Based on the exact wording of the question, yes - based on the answer choices, no.

This goes to my point of always posting the exact wording of the question - on the GMAT it will never be ambiguous.

Stuart Kovinsky wrote:

Pinku wrote:The number of words that can be formed out of the letters a,b,c,d,e,f taken 3 together each word containing atleast one vowel.
a)96 b)90 c)24 d)120 e)12

In future, please don't paraphrase questions - make sure you reproduce the exact wording. In this case we can understand what the question means, but sometimes inaccurate reproduction makes questions unanswerable. Further, please always provide the source.

On to the question!

Each word must contain at least one vowel, so there are two possible cases:

1 vowel + 2 consonants; or
2 vowels + 1 consonant.

Case 1:

2C1 * 4C2 = 2 * (4*3/2) = 2 * 6 = 12 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 12*6 = 72 possible words.

Case 2:

2C2 * 4C1 = 1 * 4 = 4 possible combinations of letters. However, we can rearrange those 3 letters in 3! = 6 ways. So, 4*6 = 24 possible words.

Either case is acceptable, so we ADD them together:

72 + 24 = 96... choose (A).

thank you.....is there any short cut method to solve it.......

Pinku wrote:The number of words that can be formed out of the letters a,b,c,d,e,f taken 3 together each word containing atleast one vowel.
a)96 b)90 c)24 d)120 e)12

total number of letters=6(a,b,c,d,e,f)

vowel=2(a,e)

consonants=4(b,c,d,f)

total number of words that can be formed is =6c3(with all letters(a,b,c,d,e,f))-4c3( without vowels(b,c,d,f))=20-4=16

number of ways=16*6=96


option A is correct