Task B-4.1. Determine the natural number $N$ for which

$$
\frac{1}{2!11!}+\frac{1}{3!10!}+\frac{1}{4!9!}+\frac{1}{5!8!}+\frac{1}{6!7!}=\frac{N}{1!12!}
$$
## Solution.

Multiply the given equation by 13!,

$$
\frac{13!}{2!11!}+\frac{13!}{3!10!}+\frac{13!}{4!9!}+\frac{13!}{5!8!}+\frac{13!}{6!7!}=\frac{N \cdot 13!}{1!12!}
$$

Then we get

$$
\binom{13}{2}+\binom{13}{3}+\binom{13}{4}+\binom{13}{5}+\binom{13}{6}=N\binom{13}{1}
$$

Add $\binom{13}{0}+\binom{13}{1}$ to both sides,

$$
\binom{13}{0}+\binom{13}{1}+\binom{13}{2}+\binom{13}{3}+\binom{13}{4}+\binom{13}{5}+\binom{13}{6}=N\binom{13}{1}+\binom{13}{0}+\binom{13}{1}
$$

The sum of all binomial coefficients in the expansion of $(a+b)^{n}$ is $2^{n}$. Then the left side of the above equation is equal to $\frac{2^{13}}{2}$ because the sum of the missing binomial coefficients, $\binom{13}{7}+\binom{13}{8}+\ldots+\binom{13}{13}$, is equal to the sum of $\binom{13}{0}+\binom{13}{1}+\ldots+\binom{13}{6}$ (due to the property of symmetry). It follows that

$$
\begin{aligned}
& \frac{2^{13}}{2}=N\binom{13}{1}+\binom{13}{0}+\binom{13}{1} \\
& 2^{12}=13 N+1+13 \\
& 2^{12}-1=13(N+1) \\
& \left(2^{6}-1\right)\left(2^{6}+1\right)=13(N+1) \\
& N=\frac{63 \cdot 65}{13}-1=314
\end{aligned}
$$