Question

Show that $C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)$.

Short answer

It is Proved that $C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)$.

Step-by-step solution

Step 1. Given Information.

Given to prove that $C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)$.

Step 2. Calculation.

When n objects are available and r are to be picked then the number of combinations is given by $C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!}$.

Consider, $C\left(n-1,r\right)+C\left(n-1,r-1\right)$

$\begin{array}{l}C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!}{\left(n-1-r\right)!r!}+\frac{\left(n-1\right)!}{\left(\left(n-1\right)-\left(r-1\right)\right)!\left(r-1\right)!}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!}{\left(n-r-1\right)!r!}+\frac{\left(n-1\right)!}{\left(n-1-r+1\right)!\left(r-1\right)!}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!\cdot \left(n-r\right)}{\left(n-r-1\right)!r!\cdot \left(n-r\right)}+\frac{\left(n-1\right)!\cdot r}{\left(n-r\right)!\left(r-1\right)!\cdot r}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!\cdot \left(n-r\right)}{\left(n-r\right)!r!}+\frac{\left(n-1\right)!\cdot r}{\left(n-r\right)!r!}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!}{\left(n-r\right)!r!}\left[n-r+r\right]\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{\left(n-1\right)!\cdot n}{\left(n-r\right)!r!}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=\frac{n!}{\left(n-r\right)!r!}\\ C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)\end{array}$

Therefore, from the values of the expressions, $C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)$.

Step 3. Conclusion.

Therefore, It is Proved that $C\left(n-1,r\right)+C\left(n-1,r-1\right)=C\left(n,r\right)$.