Problem Solving

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C(5,3)*3! = C(5,2)*6 = 10*6 = 60

STRUM 5 letters. If you are making 3 letter of UNIQUE words then use PERMUTATION as the order matters here. (Order matter means TRU is not TUR.)

So P(5,3) = 5!/(5-3)! = 60

beat_gmat_09 wrote:C(5,3)*3! = C(5,2)*6 = 10*6 = 60

If I may interject: the reason our friend multiplies the regular C(5,3) by 3! is because the order of choosing DOES matter here.

C(5,3) is the number of ways of choosing 3 letters out of 5, regardless of order. However, the question is asking for unique letter combinations: If we choose the letters S, T, U, we still have 3! letter combinations with these letters: STU is not the same as TUS, or UTS, etc.

In other words, the question is asking for permutations, not combinations: P(5,3) = 5!/(5-3)! = 5!/2! = 120/2=60.

Geva@MasterGMAT wrote:

beat_gmat_09 wrote:C(5,3)*3! = C(5,2)*6 = 10*6 = 60

If I may interject: the reason our friend multiplies the regular C(5,3) by 3! is because the order of choosing DOES matter here.

C(5,3) is the number of ways of choosing 3 letters out of 5, regardless of order. However, the question is asking for unique letter combinations: If we choose the letters S, T, U, we still have 3! letter combinations with these letters: STU is not the same as TUS, or UTS, etc.

In other words, the question is asking for permutations, not combinations: P(5,3) = 5!/(5-3)! = 5!/2! = 120/2=60.

Thank you all for your help

@ Geva: yes the 3! part was confusing. Thanks for explaining. In general I have difficulty knowing the difference between a perm and a comb, other than that it should be a piece of cake

N:Dure wrote:

Geva@MasterGMAT wrote:

beat_gmat_09 wrote:C(5,3)*3! = C(5,2)*6 = 10*6 = 60

If I may interject: the reason our friend multiplies the regular C(5,3) by 3! is because the order of choosing DOES matter here.

C(5,3) is the number of ways of choosing 3 letters out of 5, regardless of order. However, the question is asking for unique letter combinations: If we choose the letters S, T, U, we still have 3! letter combinations with these letters: STU is not the same as TUS, or UTS, etc.

In other words, the question is asking for permutations, not combinations: P(5,3) = 5!/(5-3)! = 5!/2! = 120/2=60.

Thank you all for your help

@ Geva: yes the 3! part was confusing. Thanks for explaining. In general I have difficulty knowing the difference between a perm and a comb, other than that it should be a piece of cake

that's the difference: for a permutation, the order of arrnging is important: STU is not the same as TUS.
For a combination, these two arangments, as well as all others using the letters S, T, U, are considered the same.

Therefore, the question you need to ask yourself, for each question, is "does the order matter?" For this question, is AB the same as BA, or are the two different? If the two combinations are the same (for example, for a question asking for the number of committees that can be formed, both AB and BA are the same committee - both mean that A and B were chosen, but we don't really care in which order), then the number of combinations need to be discounted by an additionak k! - which is why in C(5,3) we divide the 5!/2! by an additional 3!.

I will say this - the vast majority of GMAT questions are combinations and not permutations. In most cases, the order doesn't matter, and calculation needs to be discounted by k!| to eliminate duplicates.