# 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 $$5$$?

• Given $n+1$ people, one of which is special and the other $n$ normal, the number of ways of choosing $k$ of those people could be counted by breaking into cases based on whether or not that special person is selected. We know the total number of ways would be $\binom{n+1}{k}$. The number of ways where the special person is selected would be $\binom{n}{k-1}$ as we need $k-1$ additional normal people, and the number of ways where the special person is not selected would be $\binom{n}{k}$. This gives $\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$, Pascal's identity. – JMoravitz Nov 8 at 14:25

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 $$x-1=6-2=4$$.

• I don't see why I got $6−(x−1)=4$ while you got $6−(x−1)=2$. I'm assuming that my answer is wrong. – Axion004 Nov 8 at 14:52
• @Axion004 Note that $\binom{6}{2}=\binom{6}{4}$ therefore $x-1=2$ or $x-1=4$. – Robert Z Nov 8 at 16:21

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 $$x=5$$.

• Why does $(x-1)!$ equal to $x-1$? – Math Noob Nov 8 at 15:03
• What is inside the factorial on the LHS equals what is inside the factorial on the RHS. – Axion004 Nov 8 at 15:05
• It is worth pointing out that $(n!) = (m!)$ will imply that $n=m$ only in the case that $n,m>1$. Note that $0!=1!$ despite $0\neq 1$. – JMoravitz Nov 8 at 17:52

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!