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

Question

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

Bunuel · Accepted Answer

Manhattan GMAT Official 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. https://gmatclub.com/forum/download/file.php?id=27806 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. 2015-09-13_2041.png

kunal555 · Answer

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

Answer:-E

anudeep133 · Answer

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

shailendra79s · Answer

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

alphaseeker · Answer

Always follow the triangle, things will be easier.

Yash12345 · Answer

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).

Basshead · Answer

Really good video that explains Pascal's triangle: https://www.youtube.com/watch?v=XMriWTvPXHI

Varuag93 · Answer

You can also use the formula mentioned by  RenB  in another similar question.

When given  (x+y+z)^n , and required to find number of occurrences or sum of coefficients of  x^p*y^q*z^r -
Sum of coefficients of  x^p*y^q*z^r  =  \frac{n!}{p!*q!*r!}

In case given:  (x-y-z)^n , and required to find number of occurrences or sum of coefficients of  x^p*y^q*z^r 
Sum of coefficients of  x^p*y^q*z^r  =  \frac{n!}{p!*q!*r!}*(1)^p*(-1)^q*(-1)^r

Kinshook · Answer

Asked: In the expansion of  (x + y)^6 , what is the coefficient of the  x^3y^3  term?

(x + y)^6 = ^6C_0x^6 + ^6C_1 x^5y + ^6C_2 x^4y^2 + ^6C_3 x^3y^3 + ^6C_4 x^2y^4+ ^6C_5 xy^5 + ^6C_6y^6  
Coefficient of the  x^3y^3  term =  ^6C_3 = 20

IMO E

Adarsh_24 · Answer

Binomial expansion formula
(x + y)^n = nΣr=0 nCr x^(n – r) · y^r

We need x^3y^3
r=3
n-r=3
n=6
Coefficient nCr = 6C3=20

onlymalapink · Answer

is there an answer that d0rsnt require me to cook up a new Theorem to solve this is ridiculous

Kinshook · Answer

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

(x+y)^6 = x^6 + ^6C_1x^5y + ^6C_2x^4y^2 + ^6C_3x^3y^3 + ^6C_4x^2y^4 + ^6C_5xy^5 + y^6

The coefficient of the  x^3y^3  term =  ^6C_3 = \frac{6!}{3!3!} = 20

IMO E

DrMock · Answer

Khan academy has a great explanation of how to use pascal's triangle for binomial expansions: watch?v=v9Evg2tBdRk. (Yeah..haven't posted enough for a url apparently. Just append to the end of the yt url.)

pappal · Answer

(x+y)^6=(x+y)^3*(x+y)^3
=(x^3+y^3+3x^2*Y+3xy^2)(x^3+y^3+3x^2*y+3xy^2)
coeff. 2--from multiplying x cube of first and y cube of the other and vice versa
coeff. 18-- from multiplying 3x^2*y of first with 3xy^2 of second and the vice versa
so total coff. of x^3*y^3=20

ablatt4 · Answer

This question is hard only because it is well disguised.

This is actually a combinatorics question.

Complicated questions like this need to be broken down into peices.

(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y)

we can take 1 x and 1 y from each of these boxes so 6! options

the constraints here are we are going to take 3 x and 3y so 3!*3!

therefore this is 6!/3!*3! = 4*5=20

if this didn't make sense try and think about this as 6 boxes, in each is one x and one y, the question is telling us that we are taking one thing from each box and we need to take 3x and 3y total. The way you would solve this problem is identify everything we could potentially do (6 boxes so 6 factorial) and divide that by the constraints given (one from each 3 x 3y)

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards