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 x3y3 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 x3y3 term. You might start with a much simpler case:
(x + y) (x + y) = x2 + 2xy + y2
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 x3y3, 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 x3y3 term.
2) Use Pascal’s Triangle. This is a handy little device to get the coefficients of (x + y)n, when 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)n for some 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)n. So, for instance, you can just read off the bottom row to get all the coefficients of (x + y)6:
(x + y)6 = x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6.
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.
