Question

Prove that the value of $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$.

Answer 1

Hint: To prove $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$, we will solve L.H.S and try to show that it is equal to R.H.S. By using the formula of \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] to solve this problem.

Complete step-by-step answer:
It is given in the question to prove $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$. We know that, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] using this formula we will expand L.H.S. We get –
${{\Rightarrow }^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}$
$\begin{align}
  & \Rightarrow \dfrac{n!}{r!\left( n-r \right)!}+\dfrac{n!}{n-(r+1)!(r-1)!} \\
 & \Rightarrow \dfrac{n!}{r!\left( n-r \right)!}+\dfrac{n!}{(n-r+1)!(r-1)!} \\
\end{align}$
Taking $n!$ as common from both the terms, we get
$\Rightarrow n!\left[ \dfrac{1}{r!\left( n-r \right)!}+\dfrac{1}{(n-r+1)!(r-1)!} \right]$
We can write $r!$ as $r\times (r-1)!$ because $n!$ can be written as $n\times (n-1)\times (n-2)\times (n-3)\times ........................$
$\Rightarrow n!\left[ \dfrac{1}{r(r-1)!\left( n-r \right)!}+\dfrac{1}{(r-1)!\left( n-r+1 \right)!} \right]$
Taking $(r-1)!$ common from denominator, we get
$=\dfrac{n!}{(r-1)!}\left[ \dfrac{1}{r\left( n-r \right)!}+\dfrac{1}{\left( n-r+1 \right)!} \right]$
We can also write $(n-r+1)$ as $(n-r+1)\times (n-r)!$. Then, we get
$\Rightarrow \dfrac{n!}{(r-1)!}\left[ \dfrac{1}{(n-r)!\times r}+\dfrac{1}{(n-r+1)(n-r)!} \right]$
Taking $(n-r)!$ as common, we get
$\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{1}{r}+\dfrac{1}{(n-r+1)} \right]$
Taking L.C.M of $r$ and $(n-r+1)$ as $r(n-r+1)$, we get
$\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{n-r+1+r}{r(n-r+1)} \right]$
$\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{n+1}{r(n-r+1)} \right]$
 \[\Rightarrow \dfrac{(n+1)n!}{(r-1)!(n-r)!\times r(n-r+1)}\]
\[\Rightarrow \dfrac{(n+1)n!}{(n-r+1)(r-1)!(n-r)!\times r}\]
We can write $(n+1)\times n!$ as $(n+1)!$ because we are multiplying $n!$ to its successive number. For example, we can write $(7+1)\times 7!$ as $8!$. Also, $(n-r+1)(n-r)!$ can be written as $(n-r+1)!$ and $r(r-1)!$ can be written as $r!$.
$\Rightarrow \dfrac{(n+1)!}{(n-r+1)!\times r!}$ $={}^{n+1}{{C}_{r}}$ = L.H.S
On expanding ${}^{n+1}{{C}_{r}}$ using formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we get
${}^{n+1}{{C}_{r}}=\dfrac{(n+1)!}{(n-r+1)!\times r!}$. Therefore,
L.H.S = R.H.S.
Hence, we proved $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$.

Note: It is very important to differentiate between $^{n}{{C}_{r}}$ and ${}^{n}{{P}_{r}}$ expansion. As, $^{n}{{C}_{r}}$ gives the total combination whereas, ${}^{n}{{P}_{r}}$ gives total permutation and the expansion of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ and ${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$.