Calculators Forbidden!

We have,

${(1-{x}^{10})}^{7} = 1-\binom{7}{1}{x}^{10}+\binom{7}{2}{x}^{20}-\binom{7}{3}{x}^{30}+\binom{7}{4}{x}^{40}-\binom{7}{5}{x}^{50}+\binom{7}{6}{x}^{60}-\binom{7}{7}{x}^{70}$ ,

and,

${(1-x)}^{-7} = 1+\binom{7}{6}x+...+\binom{14}{6}{x}^{8}+...+\binom{24}{6}{x}^{18}+...+\binom{64}{6}{x}^{58}+...$

Now, if we multiply these two equations, then we can see that the above summation is nothing but the coefficients of ${x}^{58}$ in the final equation.

Thus, we need to find the coefficient of ${x}^{58}$ in ${\left ({\frac{1-{x}^{10}}{1-x}}\right )}^{7}$ .

Expanding, we need to find coefficient of ${x}^{58}$ in ${(1+x+{x}^{2}+...+{x}^9)}^{7}$ .

To solve the above question we can find the number of solutions of the equation: $x_{1}+x_{2}+...+x_{7} = 58$ , where each $0 \leq x_{i} \leq 9$

We'd like to use the formula for positive integer solutions here, but since the $x_{i}$ s are limited, it would be incorrect.

So, after a bit of manipulation, we get: $(9-x_{1})+(9-x_{2})+...+(9-x_{7}) = 5 \Rightarrow y_{1}+y_{2}+... +y_{7} = 5$ , where each $0 \leq y_{i} \leq 9$

Now the maximum allowed value of $y_{i}$ is $5$ , therefore we can simply use the formula for positive integer solution, a.k.a. Beggar's Method to get $\binom{5+7-1}{7-1} = \binom{11}{6}$

Therefore, $n-k = \fbox{5}$