Q2. - Algebra 2

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given Information.

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

Step 2. Calculation .

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

Similarly,

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

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

Step 3. Conclusion .

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