Coefficient of y^3*x^6 in (1+x+y)^5 * (1+x)^7?

Question

I actually just expanded it, and got the answer: 840.

But I`m pretty sure that there`s some binomial (tri-nomial?) expansion involved here that can lead to the answer much more efficiently.

I did try to expand the binomial using the theorem and picked the x^6 term, and then expand the trinomial, but it became a mess, because getting y^3*x^6 seems to involve more than just one term from the (1+x)^7.

All in all, can anyone shed any light on this?

Answer 1

Let B = x+y

then 1+x+y = 1+B

Pascal's triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(1+x+y)^5 = (1+B)^5

= 1^5 + 5*1^4 * B + 10 * 1^3*B^2 + 10*1^2 + B^3 + 5*1*B^4 + B^5

= 1 + 5B + 10B^2 + 10B^3 + 5B^4 + B^5

where

B^5 = (x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

b^4 = (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

B^3 = (x+y)^3 = x^3 + 3x^2 y + 3x y^2 + y^3

B^2 = (x+y)^2 = x^2 + 2xy + y^2

So the first one expands out to:

1 + 5(x+y) + 10(x^2 + 2xy + y^2) + 10(x^3 + 3x^2 y + 3x y^2 + y^3) + 5(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4)

+ x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

= 1 + 5x + 5y + 10x^2 + 20xy + 10y^2 + 10x^3 + 30x^2 y + 30xy^2 + 10 y^3 + 5x^4 + 20x^3y + 30x^2y^2 20xy^2 + 5y^4

+ x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 = Q(x,y)

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Pascal's triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

so (x+1)^7 = x^7 + 7x^6 + 21x^5 + 35x^4 + 35x^3 + 21x^2 + 7x + 1 = P(x) <--- we'll use this polynomial

we are interested in x^6 y^3, and will cross each term P(x) with select terms in Q(x,y)

to get it...

x^7 will not work.

7x^6 must be crossed with y^3 which produces 70 x^6 y^3

21x^5 must be crossed with xy^3 and there is no such term in Q(x,y)

35x^4 must be crossed with x^2y^3 which procues 350 x^6y^3

35x^3 must be crossed with x^3 y^3 which does not exist in Q(x,y)

21x^2 must be crossed with x^4 y^3 which does not exist in Q(x,y)

7x must be crossed with x^5 y^3 which does not exist in Q(x,y)

1 must be crossed with x^6 y^3 which does not exist in Q(x,y)

The total is 420 x^6 y^3