Mahtab wrote:

Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?

I understand these are all the possible combinations:

Box 1: A B

Box 2: A C

Box 3: A D

Box 4: A E

Box 5: B C

Box 6: B D

Box 7: B E

Box 8: C D

Box 9: C E

Box 10: D E

But I don't get the mathematical interpretation of this.. 5! / (3! * 2!)

I don't get how the anagram can be "YYNNN"

You can think of it in 2 ways:

1. You have 5 different types of truffles. For each box, you have to select 2 different types of truffles. In how many ways can you do it?

5C2 = 5!/3!*2!

(When out of n things, you have to select r, you can do it in nCr ways. nCr = n!/r!*(n-r)!

For more on this, check out Combinatorics & Probability by

Veritas Prep)

2. There are 5 truffles. You need to select 2 and not select the rest of the three. Y - select a truffle. N - Do not select it

So you can select 2 truffles in the number of ways in which you can arrange YYNNN.

YYNNN means you select the first 2 and leave the rest.

YNYNN means you select 1st and 3rd and leave the rest.

etc...

This can be done in 5!/2!*3!

You can arrange 5 things in 5! ways. But you have 2 Ys and 3 Ns so you divide by 2! and 3!.

(Again, for more on this, check out Combinatorics & Probability by

Veritas Prep)

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Karishma

Veritas Prep GMAT Instructor

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