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Coefficient problem.

Expert replies
by AndyB » Tue Dec 07, 2010 8:21 am
Hi All,

Could anyone please help me in understanding the explanation for the problem below:

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


(A) 6

(B) 12

(C) 15

(D) 18

(E) 20

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 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 term. You might start with a much simpler case:



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 , 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 term.




I dont understand this part -

you pick the x from the first (x + y) and the y from the second (x + y).

Regards,
Andy.
Last edited by AndyB on Tue Dec 07, 2010 9:59 am, edited 2 times in total.
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Source: — Problem Solving |

by shovan85 » Tue Dec 07, 2010 8:32 am
AndyB wrote:Hi All,

Could anyone please help me in understanding the explanation for the problem below:

In the expansion of , what is the coefficient of the term?


(A) 6

(B) 12

(C) 15

(D) 18

(E) 20

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 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 term. You might start with a much simpler case:

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 , 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 term.

I dont understand this part -

you pick the x from the first (x + y) and the y from the second (x + y).

Regards,
Andy.
Are we missing something in the question Andy? see the highlighted part in Red.

From the solution you provide I guess is it (x+y)^6 and the term is x^3 y^3 Or (xy)^3?
If the problem is Easy Respect it, if the problem is tough Attack it
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by AndyB » Tue Dec 07, 2010 10:00 am
Yes Shovan , You are rite...
Have updated the question with the correct info now.
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by Rahul@gurome » Tue Dec 07, 2010 10:42 am
AndyB wrote:In the expansion of (x + y)^6, what is the coefficient of the term x^3 * y^3?
(A) 6
(B) 12
(C) 15
(D) 18
(E) 20
Observe that x³y³ is nothing but a grouping of 3 'x's and 3 'y' s.
Some of such possible groupings are: (xxxyyy), (xxyyxy), (xyxyxy), (yyxyxx) etc.

Now coefficient of x³y³ in the expansion of (x + y)^6 is nothing but the total number of occurrences of such groups in the expansion. Which is again same as the number of possible arrangements among 3 'x's and 3 'y's = 6!/(3!*3!) = 20

The correct answer is E.
Rahul Lakhani
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