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

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In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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07 Sep 2015, 04:38
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In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

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Posts: 56357
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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13 Sep 2015, 09:42
1
1
4
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

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).

Attachment:

2015-09-13_2041.png [ 4.28 KiB | Viewed 70268 times ]

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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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07 Sep 2015, 12:49
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5
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

Method 1:-
By using traditional algebraic epansion we can find that coefficient of $$x^3 y^3$$ is 20

Method 2:-
By using pascal's triangle,we can find out coefficient of all the terms in the expression.
the power of the expression is 6.
so there will be 7 terms in the expansion and hence the 7th row of pascal's triangle will be considered i.e
1 6 15 20 15 6 1
so the coefficient will be 20
Verification with expansion
$$(x+y)^6 = x^6 +6 x^5 y + 15x^4 y^2 +20x^3 y^3 +15x^2 y^4 +6xy^5 + y^6$$

Method 3:-
By binomial expansion, we can directly find out the term.
$$6_C_3$$$$*x^{6-3}*y^3$$$$=20x^3 y^3$$

##### General Discussion
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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07 Sep 2015, 07:13
2
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
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In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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07 Sep 2015, 15:12
As mentioned before, I just wrote out Pascal's triangle quickly to get the answer of 20.
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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07 Sep 2015, 22:23
1
(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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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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09 Sep 2015, 11:06
2
Always follow the triangle, things will be easier.
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Please kindly click on "+1 Kudos", if you think my post is useful
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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09 Sep 2015, 14:37
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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13 Sep 2015, 10:13
1
1
Bunuel wrote:
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

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).

Attachment:
2015-09-13_2041.png

Is the methodology right to derive the answer. ?
I derived it this way

(x+y)^6
=(x+y)^3 * (x+y)^3
=(x^3+Y^3+3x^2*y+3y^2*x) (x^3+Y^3+3x^2*y+3y^2*x)
Now since we want the ans in x^3 Y^3 terms , we will only multiply those terms in left bracket to those terms in right bracket which yield x^3 Y^3.

For eg . 3x^2*y(from left bracket ) multiply to 3y^2*x ( from right bracket)= 9x^3*y^3
similarly 3y^2*x (from left bracket ) multiply to 3x^2*y ( from right bracket)= 9x^3*y^3
and (now single terms): x^3(from left bracket ) multiply to y^3 ( from right bracket) = x^3*y^3
similarly, y^3(from left bracket ) multiply to x^3( from right bracket)= x^3*y^3

Adding all these values we get = 20 *x^3 * Y^3 .. Thus ans (E).
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In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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14 Feb 2016, 15:47
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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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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25 Feb 2016, 23:57
Banual,

I have one query these methods whether the pascals or combination as 6|3|*3| will hold good for equations as this (a-b)^6 ???
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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02 May 2017, 00:12
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
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

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05 Jul 2019, 06:55
Bunuel wrote:
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

Kudos for a correct solution.

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).

Attachment:
2015-09-13_2041.png

Bunuel

I have seen some Binomial expansion solutions, so just wanted to make sure whether binomial expansion comes under the scope of GMAT or not.
So could you please confirm whether we can expect questions which needs a binomial expansion for the solution.

Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3   [#permalink] 05 Jul 2019, 06:55
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