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 $5$?
 A: 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 $x-1=6-2=4$.
A: 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 $x=5$.
A: The formula can be edited to reason the answer, Here's How
$^6C_2+^6C_X=^7C_X$ implies $^6C_2=^7C_X-^6C_X$    (1)
We have to find $x$
we have $^nC_{(k-1)}+^nC_k=^{(n+1)}C_k$
Put $n=6$ and $k=x$
6C(x-1)+6Cx=7Cx
6C(x-1)=7Cx-6Cx (2)  
COMPARING 1 and 2
6C(x-1)=6C2
implying x=3 or x=5
Though it may appear that the answer can't be achieved through a given formula BUT with a few gimmicks anything can happen!
