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Let $k[x]=k[x_1,\dots,x_n]$ be the set of polynomials in $n$ variables. Given a $k<n$, how can you determine the dimension of the vector subspace of polynomials of degree $\le k$?

I guessed it would be $\frac{k(k+1)\cdots(k+n-1)}{n!}$, with the $n!$ since you have various choices of bases. I am having trouble justifying this intuition, or even confirming if it is correct.

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up vote 2 down vote accepted

You are close, but you are only considering the terms of degree exactly $k$, not $k$ or less.

Consider the standard basis you would choose. The terms of the polynomial have the form $$x_1^{k_1}\cdot x_2^{k_2} \cdots {x_n}^{k_n}$$ where $k_1 + \cdots + k_n \le k$. Therefore you are counting the number of solutions to $$k_1 + \cdots + k_n = i$$ for $0\le i \le k$. Summing over the range of $i$, the dimension is given by $$\sum_{i=0}^k\binom{n+i-1}{i}=\binom{n+k}{n}$$

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