Question 1137744

determine the coefficient of x^9 y^4 in the binomial expansion of (x + y)^13 power
<pre>In the BINOMIAL EXPANSION of {{{(x + y)^13}}}, x<sup>9</sup>y<sup>9</sup> occurs at the term where: {{{matrix(1,3, x^9y^4, "=", " " [13]C[r-1] * (x)^(13-(r-1))(y)^(r-1))}}}
As such, 9 = 13  -  (r  -  1)              OR                4 = r  -  1
9 = 13  -  r + 1                                             5 = r
9  -  14 = - r
- 5 = - r
5 = r
With the coefficients on x and y in {{{(x + y)^13}}} being 1, the coefficient on the 5th term, when expanded, will be the value in the 5th COLUMN of the 13th ROW of Pascal’s triangle, which is {{{highlight_green(715)}}}
OR
With the coefficient on x and y being 1, the coefficient on the 5th term will be: {{{highlight_green(system(matrix(1,9, " "[n]C[r-1], where, n, "=", 13, and, r, "=", 5), matrix(1,7, " " [n]C[r-1], becomes, " "[13]C[5-1], "=", " " [13]C[4], "=", 715)))}}}
<b>FYI: The entire EXPANSION DOESN'T HAVE to be done, as one person decided to do!</b>
What if {{{x^9y^4}}} was the 24th or 25th term of the expansion? Do you think it'd be wise to do the entire expansion? Think about it!