- #1

- 1,015

- 81

- Homework Statement:
- "Prove that the coefficient for ##x^ky^{n-k}## is ##\binom n k##."

- Relevant Equations:
- Derived expression for binomial coefficient: ##\sum_{m=1}^{n-k+1} \sum_{i=1}^m i##

Basically, the way I did this problem was to use a table with a known ##n## and ##k##. In this case, I fixed ##n=5##, and ##k=3##. I wanted to find the number of terms with three ##x##'s and two ##y##'s. I labeled each ##x_i##, ##1\leq i \leq 5##; the ##y_i## are labeled the same way. Anyway, what I did was create a table that helped me keep count of how many possible terms I can have with exactly three ##x##'s. It looks something like this: I made a matrix (table) with the indices at the top, the ##k-1## variables I fix as ##X##, the possible choices for the last ##x_i## as ##O##, and the variable(s) I exclude as ##+##.

##\begin{pmatrix}

1 & 2 & 3 & 4 & 5\\

X & X & O & O & O \\

X & + & X & O & O \\

X & + & + & X & O\\

+ & X & X & O & O \\

+ & X & + & X & O \\

+ & + & X & X & O \\

\end{pmatrix}##

What I do is count the number of ##O##'s there are to get the number of terms with exactly ##k## ##x_i## in them. I get that this is not exactly orthodox, but that is how I got the expression in the relevant equations section. The trouble I have is, that I am not sure that it is completely correct, and that I am having trouble equating it to ##\binom n k##. I imagine it will take a lot of strenuous algebra, which is why I want to hold off on it until someone can confirm my suspicions that:

##\binom n k = \sum_{m=1}^{n-k+1} \sum_{i=1}^m i##

##\begin{pmatrix}

1 & 2 & 3 & 4 & 5\\

X & X & O & O & O \\

X & + & X & O & O \\

X & + & + & X & O\\

+ & X & X & O & O \\

+ & X & + & X & O \\

+ & + & X & X & O \\

\end{pmatrix}##

What I do is count the number of ##O##'s there are to get the number of terms with exactly ##k## ##x_i## in them. I get that this is not exactly orthodox, but that is how I got the expression in the relevant equations section. The trouble I have is, that I am not sure that it is completely correct, and that I am having trouble equating it to ##\binom n k##. I imagine it will take a lot of strenuous algebra, which is why I want to hold off on it until someone can confirm my suspicions that:

##\binom n k = \sum_{m=1}^{n-k+1} \sum_{i=1}^m i##