Question

${}^{n}C_r+{}^{n}C_{r-1}$ is equal to?

• A. ${}^{n+1}C_r$
• B. ${}^{n}C_{r+1}$
• C. ${}^{n+1}C_{r+1}$
• D. ${}^{n-1}C_{r-1}$

Answer 1

The correct option is A

${}^{n+1}C_r$

Explanation for the correct answer:

Expanding the given expression to find the value:

Given expression is ${}^{n}C_r+{}^{n}C_{r-1}$

Now, expanding the given expression we get,

$=\frac{n!}{r!\left(n-r\right)!}+\frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}\left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$=\frac{n!}{r\left(r-1\right)!\left(n-r\right)!}+\frac{n!}{\left(r-1\right)!\left(n-r+1\right)\left(n-r\right)!}$

$=\frac{n!}{\left(r-1\right)!\left(n-r\right)!}\left[\frac{1}{r}+\frac{1}{n-r+1}\right]$

$=\frac{n!}{\left(r-1\right)!\left(n-r\right)!}\left[\frac{n+1}{r\left(n-r+1\right)}\right]$

$=\frac{\left(n+1\right)!}{\left(r\right)!\left(n-r+1\right)!}\left[\text{this expression is of the format}\frac{n!}{\left(r\right)!\left(n-r\right)!}\right]$

$={}^{n+1}C_r$

Hence, option (A) is the correct answer.