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^3y^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^3y^3 term. You might start with a much simpler case:

(x + y) (x + y) = x^2 + 2xy + 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^3y^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^3y^3 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.



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:

\((x + y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\).

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.


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Solution:

Trick is we have to eliminate terms with powers wherever we see them.
\((x+y)^6 = (x+y)^2*(x+y)^2*(x+y)^2 = (x^2 + y^2 + 2xy)*(x^2 + y^2 + 2xy)*(x+y)^2\)
= \((6 x^2 y^2 + 4 x^3 y + 4 x y^3)*(x^2 + y^2 + 2xy)\) (eliminated higher power terms)
= \(4 x^3 y^3 + 4 x^3 y^3 + 12 x^3 y^3\)
= \(20 x^3 y^3\)

So, Option E

(x+y)^6 = (x+y)^2 * (x+y)^2 * (x+y)^2
By expanding coefficient of x^3*y^3 = 20
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Answer E

procedure to follow pascal triangle.

the question is a nightmare...took 4 minutes to get to the answer choice...expanded everything....
where can I read about the pascal's triangle and ways to use it?
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We can use binomial theorem here.

(x+y)^n = nC0 x^ny^0 + nc1 x^(n-1)y^1 + .....nCn x^0 * y^n
(a + b)5 = 5C0a5 + 5C1a4b + 5C2a3b2 + 5C3a2b3 + 5C4ab4 + 5C5b5