How many numbers?

@Melissa Quail , actual is it possible to solve it without using Wolfram Alpha. As you have say in the comment below the answer is the coefficient of $x^{25}$ in this expansion of $(1+x+...+x^9)^6$ . Now:

$(1+x+x^2+...+x^9)^6=\frac{(1-x^10)^6}{(1-x)^6}$ and so, using Binomial Theorem we could write:

$(1-x^{10})^6=\sum_{k=0}^{6}{{6}\choose{k}}(-1)^{k}x^{10k}$ and using Negative Binomial Theorem $(1-x)^{-6}=\sum_{s=0}^{\infty}{{5+s}\choose{s}}x^{6}$

Putting all together:

$(1+x+x^2+...x^9)^6= \sum_{s=0}^{\infty}\sum_{k=0}^{6}{{6}\choose{k}}{{5+s}\choose{s}}(-1)^{k}x^{10k+s}$

Now $10k+s=25$ when $(k,s)=(0,25)\wedge(1,15)\wedge(2,5)$ and so the coefficient is:

${{6}\choose{0}}{{30}\choose{25}}-{{6}\choose{1}}{{20}\choose{15}}+{{6}\choose{2}}{{10}\choose{5}}=53262$ .

I added a summation to the formula given by Garrett Clarke for $k \ = \ 3 \ to \ 7$ in this problem . Obviously, I generated a loop summation for it in python. How did you do it? What's the mathematical approach?