*Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the 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 *x*^{3}*y*^{3} 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

*x*^{3}*y*^{3}term. You might start with a much simpler case:(

*x*+*y*) (*x*+*y*) =*x*^{2}+ 2*xy*+*y*^{2}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*x*^{3}*y*^{3}, 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*x*^{3}*y*^{3}term.2) Use Pascal’s Triangle. This is a handy little device to get the coefficients of (

*x*+*y*)*, when*^{n}*n*is a relatively small integer. You build the triangle downwards with 1’s on the outside. All the interior numbers are sums of the two numbers above.1 | ||||||||||||

1 | 1 | |||||||||||

1 | 2 | 1 | ||||||||||

1 | 3 | 3 | 1 | |||||||||

1 | 4 | 6 | 4 | 1 | ||||||||

1 | 5 | 10 | 10 | 5 | 1 | |||||||

1 | 6 | 15 | 20 | 15 | 6 | 1 |

Each row gives you the coefficients of (

*x*+*y*)*for some*^{n}*n*. Since the*second*row gives you 1 and 1 (the coefficients of (*x*+*y*)^{1}=*x*+*y*, it’s actually the*n+1*’th row that gives you the coefficients of (*x*+*y*)*. So, for instance, you can just read off the bottom row to get all the coefficients of (*^{n}*x*+*y*)^{6}:(

*x*+*y*)^{6}=*x*^{6}+ 6*x*^{5}*y*+ 15*x*^{4}*y*^{2}+ 20*x*^{3}*y*^{3}+ 15*x*^{2}*y*^{4}+ 6*xy*^{5}+*y*^{6}.The reason this works is because each number in Pascal’s triangle represents the number of legal “zigzags” that get you to that number from the 1 at the top. (A legal zigzag goes down one row and right or left just one number.) For instance, there are 20 legal zigzags that go from the 1 at the top of the triangle to the number 20 in the middle. Each of those zigzags has 3 left “zigs” and 3 right “zags” in it. For instance, you could go left-left-left-right-right-right and end up at 20, or you could go left-right-left-right-left-right and end up at 20, etc. This is exactly the same situation as counting the anagrams of a 6-letter word with two repeated letters (x and y, or L and R).

**The correct answer is E.**