Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

A) 5C2

B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)

C) - 5C3

D) 5C4

More than one options may be correct.

- Feb 13th 2009, 10:27 PMfardeen_genCoefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is?
Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

A) 5C2

B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)

C) - 5C3

D) 5C4

More than one options may be correct. - Feb 13th 2009, 10:50 PMred_dog
The general term of is

The general term of is

Then the genaral term of the entire expansion is

So the condition is .

Now take all the possibilities for k and l. - Feb 13th 2009, 10:55 PMDeMath
- Feb 13th 2009, 11:08 PMfardeen_genQuote:

The general term of http://www.mathhelpforum.com/math-he...28391dc6-1.gif is http://www.mathhelpforum.com/math-he...53b23532-1.gif

The general term of http://www.mathhelpforum.com/math-he...021d471d-1.gif is http://www.mathhelpforum.com/math-he...7202a132-1.gif

Then the genaral term of the entire expansion is http://www.mathhelpforum.com/math-he...d22b7052-1.gif

So the condition is http://www.mathhelpforum.com/math-he...9059a502-1.gif.

Now take all the possibilities for k and l.

- Feb 13th 2009, 11:23 PMred_dog
No.

You have the possibilities:

k=0, l=4

k=1, l=3

k=2, l=2

k=3, l=1

k=4, l=0

Then the coefficient is

- Feb 13th 2009, 11:29 PMDeMath
Members of the decomposition of the factor have only even powers. Therefore, you just enough to consider this factor.

Your answer must be for . - Feb 13th 2009, 11:37 PMfardeen_gen
So correct answers are:

B), C) and D), right? - Feb 13th 2009, 11:43 PMred_dog
No, the correct answer is B.

- Feb 13th 2009, 11:47 PMDeMath

Sorry, I corrected. - Feb 13th 2009, 11:53 PMred_dog
- Feb 13th 2009, 11:56 PMfardeen_gen
red_dog, A) is equivalent to B). So A) is also correct, right?

And aren't we supposed to take the individual possibility of k = 4 and l = 0, which gives 5C4(option D)?