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Question: Prove that the value of \(^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\)....

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

Explanation

Solution

Hint: To prove nCr+nCr−1=n+1Cr^{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 nCr=n!r!(n−r)!{}^{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 nCr+nCr−1=n+1Cr^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}. We know that, nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} using this formula we will expand L.H.S. We get –
⇒nCr+nCr−1{{\Rightarrow }^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}
⇒n!r!(n−r)!+n!n−(r+1)!(r−1)! ⇒n!r!(n−r)!+n!(n−r+1)!(r−1)! \begin{aligned} & \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{aligned}
Taking n!n! as common from both the terms, we get
⇒n![1r!(n−r)!+1(n−r+1)!(r−1)!]\Rightarrow n!\left[ \dfrac{1}{r!\left( n-r \right)!}+\dfrac{1}{(n-r+1)!(r-1)!} \right]
We can write r!r! as r×(r−1)!r\times (r-1)! because n!n! can be written as n×(n−1)×(n−2)×(n−3)×........................n\times (n-1)\times (n-2)\times (n-3)\times ........................
⇒n![1r(r−1)!(n−r)!+1(r−1)!(n−r+1)!]\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)!(r-1)! common from denominator, we get
=n!(r−1)![1r(n−r)!+1(n−r+1)!]=\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)(n-r+1) as (n−r+1)×(n−r)!(n-r+1)\times (n-r)!. Then, we get
⇒n!(r−1)![1(n−r)!×r+1(n−r+1)(n−r)!]\Rightarrow \dfrac{n!}{(r-1)!}\left[ \dfrac{1}{(n-r)!\times r}+\dfrac{1}{(n-r+1)(n-r)!} \right]
Taking (n−r)!(n-r)! as common, we get
⇒n!(n−r)!(r−1)![1r+1(n−r+1)]\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{1}{r}+\dfrac{1}{(n-r+1)} \right]
Taking L.C.M of rr and (n−r+1)(n-r+1) as r(n−r+1)r(n-r+1), we get
⇒n!(n−r)!(r−1)![n−r+1+rr(n−r+1)]\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{n-r+1+r}{r(n-r+1)} \right]
⇒n!(n−r)!(r−1)![n+1r(n−r+1)]\Rightarrow \dfrac{n!}{(n-r)!(r-1)!}\left[ \dfrac{n+1}{r(n-r+1)} \right]
⇒(n+1)n!(r−1)!(n−r)!×r(n−r+1)\Rightarrow \dfrac{(n+1)n!}{(r-1)!(n-r)!\times r(n-r+1)}
⇒(n+1)n!(n−r+1)(r−1)!(n−r)!×r\Rightarrow \dfrac{(n+1)n!}{(n-r+1)(r-1)!(n-r)!\times r}
We can write (n+1)×n!(n+1)\times n! as (n+1)!(n+1)! because we are multiplying n!n! to its successive number. For example, we can write (7+1)×7!(7+1)\times 7! as 8!8!. Also, (n−r+1)(n−r)!(n-r+1)(n-r)! can be written as (n−r+1)!(n-r+1)! and r(r−1)!r(r-1)! can be written as r!r!.
⇒(n+1)!(n−r+1)!×r!\Rightarrow \dfrac{(n+1)!}{(n-r+1)!\times r!} =n+1Cr={}^{n+1}{{C}_{r}} = L.H.S
On expanding n+1Cr{}^{n+1}{{C}_{r}} using formula nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}, we get
n+1Cr=(n+1)!(n−r+1)!×r!{}^{n+1}{{C}_{r}}=\dfrac{(n+1)!}{(n-r+1)!\times r!}. Therefore,
L.H.S = R.H.S.
Hence, we proved nCr+nCr−1=n+1Cr^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}={}^{n+1}{{C}_{r}}.

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