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 $5$?

  • $\begingroup$ 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. $\endgroup$ – 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 $x-1=6-2=4$.

  • $\begingroup$ 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. $\endgroup$ – Axion004 Nov 8 at 14:52
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    $\begingroup$ @Axion004 Note that $\binom{6}{2}=\binom{6}{4}$ therefore $x-1=2$ or $x-1=4$. $\endgroup$ – 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

$$\begin{pmatrix} 6\\4 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{4!2!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$ $$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$


$$(x-1)! = 4! \implies x-1=4 \implies x=5$$ $$(6-(x-1))!=2! \implies 6 -(x-1)=2 \implies -x=-4-1\implies x=5$$ which gives us our second solution of $x=5$.

  • $\begingroup$ Why does $(x-1)!$ equal to $x-1$? $\endgroup$ – Math Noob Nov 8 at 15:03
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    $\begingroup$ What is inside the factorial on the LHS equals what is inside the factorial on the RHS. $\endgroup$ – Axion004 Nov 8 at 15:05
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    $\begingroup$ 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 $0\neq 1$. $\endgroup$ – 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)=7Cx-6Cx (2)


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!


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