Question

Prove $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $

Answer 1

Hint: To solve this question we will use the concept of combination. We will first consider the LHS of the given equation by using the formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] . Then we will expand the LHS by substituting the formula and solve further to show that it is equal to RHS.

Complete step by step answer:
We have been given an expression $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ .
Let us first consider the LHS of the equation we have $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}} $
Now, we know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Substituting the value in the given equation we get
\[\Rightarrow \dfrac{n!}{r!\left( n-r \right)!}+\dfrac{n!}{\left( r-1 \right)!\left( n-\left( r-1 \right) \right)!}\]
Now we know that $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)!.... $
Now, solving further we get
\[\Rightarrow \dfrac{n!}{r\left( r-1 \right)!\left( n-r \right)!}+\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)\left( n-r \right)!}\]
Now, taking the common terms out, we get
\[\Rightarrow \dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{1}{\left( n-r+1 \right)}+\dfrac{1}{r} \right]\]
Now, taking LCM and solving further we get
\[\Rightarrow \dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{r+n-r+1}{r\left( n-r+1 \right)} \right]\]
Now, simplifying further we get
\[\Rightarrow \dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{n+1}{r\left( n-r+1 \right)} \right]\]
\[\Rightarrow \dfrac{n!\left( n+1 \right)}{\left( r-1 \right)!\left( n-r \right)!r\left( n-r+1 \right)}\]
Now, we can write the above equation as \[\Rightarrow \dfrac{\left( n+1 \right)!}{r!\left( n+1-r \right)!}\]
When we compare the obtained equation with the standard formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] we observe that it is equal to $ {}^{n+1}{{C}_{r}} $ .
So, by solving the LHS we get $ {}^{n+1}{{C}_{r}} $ .
Which is equal to the RHS $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ .
Hence proved

Note:
 The key concept to solve this type of question is to first try to expand the LHS and get the value equal to the RHS. In this question opening of $ n! $ is not necessary, it complicates the question. Avoid common calculation mistakes. Please keep in mind the definition and formula of combination. The combination is the number of possible arrangements in a collection of items where the order of selection doesn’t matter.