Solve binomial coefficient equation My book asks me to solve this equation:
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\begin{pmatrix} 7\\x \end{pmatrix}$$
The solution is $x=3$ and the formula $$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n\\k \end{pmatrix}=\begin{pmatrix} n+1\\k \end{pmatrix}$$ is supposed to reason that solution. What I do not understand however is, if $n=6$ would $n-1$ not equal $5$?

The formula should be
$$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}.$$
It is the fundamental recurrence of the binomial coefficients.
Hence
$$\binom{6}{2}+\binom{6}{x}=\binom{7}{x}=\binom{6}{x-1}+\binom{6}{x}\implies \binom{6}{2}=
\binom{6}{x-1}.$$
and, by symmetry, it follows that we have TWO solutions: $x-1=2$ OR $x-1=6-2=4$.
