Find the coefficient of x^4y^3 in the expansion of (x+2y)^7

Answer 1

Review of binomial expansion:
http://www.purplemath.com/modules/binomial.htm

(a+b)^n
=a^n+na^(n-1)b+(n(n-1)/2)a^(n-2)b²+...+nab^(n-1)+b^n
=C(n,0)a^n+C(n,1)a^(n-1)b+C(n-2)a^(n-2)b^sup2;.....C(n,n)b^n

For n=1
(a+b)¹=a+b
(a+b)²=a²+2ab+b^sup2;
(a+b)³=a³+3a^sup2;b+3ab²+b³
...
The coefficients are therefore
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
...
This is the Pascal's triangle
http://www.mathsisfun.com/pascals-triangle.html
the sum of each pair of adjacent numbers gives rise to a number on the next line.

This also means that
(a+b)⁷
=a^7+7a^6b+21a^5b²+35a^4b³+....+7ab^6+b^7 ............(A)
which is almost the solution to the problem.
To solve the problem using the Pascal's triangle, substitute a=x, and b=2y in equation (A) to give the expression of the final answer.

I strongly suggest you review the subject on binomial expansion and Pascal's triangle before proceeding to solve the problem. If you do not have a textbook, use the links given above.