Permutation of subset of a set with repeated elements

[ronaldramlan]

Hi,

Can anyone help me on how to calculate the permutation of a subset with repeated elements?

For instance, if one needs to choose 3 letters out of 7-letter set {A,A,A,B,B,C,D}, how many ways (different positions count) can one have?

Thanks ..

Regards,
Ronald

[bachiter]

For instance, if one needs to choose 3 letters out of 7-letter set {A,A,A,B,B,C,D}, how many ways (different positions count) can one have?


It will be C(7,3)/(3!*2!) this is if u have to choose.
If u have to arrange the it will be 7!/(3!*2!).

[shiven.bhandari]

I believe it is

P (7, 3)/(3! * 2!) because the order counts.

May be an expert could shed more light..

[ronaldramlan]

Thanks, guys.
Originally, I thought that either C(7,3)/(3!*2!) or P(7,3)/(3!*2!) would work, but then I realized that none of them results in an integer, an impossible outcome for such question.

[mschwrtz]

Hmm... I don't think that you'll ever see anything quite like this on the GMAT, but here you go,

Consider how many letters the words might have in common: three, XXX; two, XXY; or none, XYZ. For each group of three common letters, there is one "word." For each group of two common letters, there are three (3!/2!) "words," XXY, XYX, and YXX. For each group with no common letters there are six (3!) "words," XYZ, XZY, YXZ, YZX, ZXY, and ZYX.

All that remains is to figure out how many groups there are of each type. There is one group of type XXX (AAA); there are six groups of type XXY (AA and any of B, C, D, and BB and any of A, C, D); and there are four groups of type XYZ (4!/3!, or just four different letters any one of which could be left out).

So...

1 group of type XXX times 1 "word" per group = 1 word.
6 group of type XXY times 3 "words" per group = 18 word.
4 group of type XYZ times 4 "words" per group = 24 word.

43 "word" all together.