GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 08 Dec 2019, 04:00 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the expansion of (x + y)^6, what is the coefficient of the x^3y^3

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59592
In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

2
16 00:00

Difficulty:

(N/A)

Question Stats: 47% (01:42) correct 53% (02:02) wrong based on 276 sessions

### HideShow timer Statistics

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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59592
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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 73188 times ]

_________________
Manager  Joined: 29 Jul 2015
Posts: 152
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

7
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
Manager  Joined: 10 Aug 2015
Posts: 102
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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
Intern  Joined: 02 Jun 2015
Posts: 32
Location: United States
Concentration: Operations, Technology
GMAT Date: 08-22-2015
GPA: 3.92
WE: Science (Other)
In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

As mentioned before, I just wrote out Pascal's triangle quickly to get the answer of 20.
Intern  Joined: 03 Feb 2014
Posts: 37
Location: United States
Concentration: Entrepreneurship, General Management
WE: General Management (Other)
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

1
(x+y)^6 = (x+y)^2 * (x+y)^2 * (x+y)^2
By expanding coefficient of x^3*y^3 = 20
_________________
--Shailendra
Manager  Joined: 01 Jun 2013
Posts: 101
GMAT 1: 650 Q50 V27 Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

2
Always follow the triangle, things will be easier.
Attachments Algebra.PNG [ 4.92 KiB | Viewed 53597 times ]

Senior Manager  P
Joined: 27 Aug 2014
Posts: 368
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
GPA: 3.9
WE: Analyst (Energy and Utilities)
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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.

Intern  Joined: 21 Sep 2013
Posts: 26
Location: United States
Concentration: Finance, General Management
GMAT Date: 10-25-2013
GPA: 3
WE: Operations (Mutual Funds and Brokerage)
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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).
Board of Directors P
Joined: 17 Jul 2014
Posts: 2492
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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?
Manager  Joined: 03 May 2013
Posts: 68
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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 ???
Manager  S
Joined: 17 Aug 2012
Posts: 119
Location: India
Concentration: General Management, Strategy
Schools: Copenhagen, ESMT"19
GPA: 3.75
WE: Consulting (Energy and Utilities)
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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
Intern  B
Joined: 03 Jun 2019
Posts: 22
Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3  [#permalink]

### Show Tags

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
Display posts from previous: Sort by

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