A ball can be colored with any one of the following $K$ colors:
$\{c_{1}, c_{2}, c_{3}, \dots, c_{k}\}$
A box is filled with balls, such that there are $10$ balls of each color, thereby containing a total of $10.K$ balls.
$\{\\
b_{c_1}^{1}, b_{c_1}^{2}, b_{c_1}^{3} \dots, b_{c_1}^{10},\\
b_{c_2}^{1}, b_{c_2}^{2}, b_{c2}^{3} \dots, b_{c_2}^{10},\\
\vdots\\
b_{c_k}^{1}, b_{c_k}^{2}, b_{c_k}^{3} \dots, b_{c_k}^{10}\\\}= 10.K$
All the balls having same color are totally identical.
In how many ways can we select $N$ balls from the box? $(N<10K)$
Suppose we had only $4$ colors, $a,b,c,d$. Represent the choice of $n_a$ balls of color $a$, $n_b$ of color $b$, $n_c$ of color $c$ and $n_d$ of color $d$ by the term $$a^{n_a}b^{n_b}c^{n_c}d^{n_d}$$ Consider the product $$(1+a+\cdots+a^{10})(1+b+\cdots+b^{10})(1+c+\cdots+c^{10})(1+d+\cdots+d^{10})$$ If we count the numbers of terms $a^{n_a}b^{n_b}c^{n_c}d^{n_d}$ with $a_n+b_n+c_n+d_n=N$, that gives the number of ways to choose $N$ balls, since we must choose from $0$ to $10$ of color $a$, and so on for each color. Now, set $a=b=c=d=x$ and we see that the number is just the coefficient of $x^N$ in $$(1+x+\cdots+x^{10})^4.$$
The same kind of argument works for any number of colors. Now to figure out what that coefficient is, write $$(1+x+\cdots+x^{10})^K=\left({{1-x^{11}}\over{1-x}}\right)^K=(1-x^{11})^K(1-x)^{-K}$$ and expand the first term by the binomial formula, and the second by the binomial series. Then for each term of degree $d\leq N$ in the first factor, you just have to multiply the coefficient by the coefficient of the term of degree $N-d$ in the second formula, and add up all the products.