Coefficient problem.

Hi All,

Could anyone please help me in understanding the explanation for the problem below:

In the expansion of (x+y)^6, what is the coefficient of the term x^3 * y^3?


(A) 6

(B) 12

(C) 15

(D) 18

(E) 20

Answer
There is a long way to solve this problem, of course: write out (x + y) (x + y) (x + y) (x + y) (x + y) (x + y), expand mechanically, and get the coefficient of the term. This would take too long on the GMAT, however. There are at least 2 shortcuts.

1) Start mechanically, but think about what you’re doing to make the term. You might start with a much simpler case:



Notice that you get a 2 on the xy term, because there are two xy products you can form as you expand:

(X + y) (x + Y) – you pick the x from the first (x + y) and the y from the second (x + y).

(x + Y) (X + y) – vice versa.

If you want to expand a much bigger product of (x + y)’s and find the coefficient of a particular term such as , then you need to think about all the different ways you can get three x’s and three y’s as you expand.

(X + y) (X + y) (X + y) (x + Y) (x + Y) (x + Y) – pick the three x’s first, then the three y’s.

(X + y) (x + Y) (X + y) (x + Y) (X + y) (x + Y) – pick x, y, x, y, x, y. etc.

So really what you’re asking is this: how many ways can you rearrange three x’s and three y’s! That’s a combinatorics problem (how many anagrams are there of the word “xxxyyy”?). The number of ways to rearrange these letters is 6!/(3!3!) = 20, and 20 is the coefficient on the term.




I dont understand this part -

you pick the x from the first (x + y) and the y from the second (x + y).

Regards,
Andy.