Integer solutions of the following question:

Question

How many integer solutions are there in this equation:

$$x_1 + x_2 + x_3 + x_4 + x_5 = 63, \quad x_i \ge 0, \quad x_2 ≥10$$

I got $C(56,3)$.

Is that correct?

Answer 1

Give $10$ immediately to $x_2$ (maybe more later) leaving $k=63-10=53$ to be distributed. Now it is a standard $\binom{n+k-1}k$ problem, with $n=5$. The answer is $\binom{57}{53}=\binom{57}4=395010$.

Answer 2

No, give 10 things to $x_2$ 53 things left , any 5 of them can get any value so answer is $\frac{57!}{4!53!}$. Basically formula for distribution of $n$ identical things in $r$ distinct bins where each bin can get any number of things is $\frac{(n+r-1)!}{n!(r-1)!}$

Answer 3

Substitute $y_1 = x_1 + 1, y_2 = x_2 - 9, y_3 = x_3 + 1, y_4 = x_4 + 1, y_5 = x_5 + 1$. This yields the equivalent problem $$ (y_1 - 1) + (y_2 + 9) + (y_3 - 1) + (y_4 - 1) + (y_5 - 1) = 63; y_i \ge 1 $$ i.e. $$ y_1 + y_2 + y_3 + \cdots + y_5 = 58 $$ By a classic stars-and-bars argument (writing $58$ in unary) the answer is then $$ {58 - 1 \choose 5 - 1} = {57 \choose 4}. $$