<s>[INST] 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 $5$?
 [/INST] We have
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{2!4!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
$$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
therefore since $(n)!=(m)!$ implies $n=m$ only when $n,m > 1$ we see that 
$$(x-1)! = 2! \implies x-1=2 \implies x=3$$
$$(6-(x-1))!=4! \implies 6 -(x-1)=4 \implies -x=-2-1\implies x=3$$
which gives us our first solution of $x=3$. Next, because
$$\begin{pmatrix} 6\\2 \end{pmatrix}=\begin{pmatrix} 6\\4 \end{pmatrix}$$
we have
$$\begin{pmatrix} 6\\4 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{4!2!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
$$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
hence
$$(x-1)! = 4! \implies x-1=4 \implies x=5$$
$$(6-(x-1))!=2! \implies 6 -(x-1)=2 \implies -x=-4-1\implies x=5$$
which gives us our second solution of $x=5$.
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