Chapter 0, Problem 16PO

To determine

To calculate: Number of ways to select one competitor and one alternate out of 8 students.

Answer to Problem 16PO

There are 56 ways.

Explanation of Solution

Given information:

  $n_1=8$ and $n_2=2$

  $r_1=2$ and $r_2=1$

Used formula: $\left(\frac{{{\displaystyle n}}_1!}{{{\displaystyle (n}}_1-{{\displaystyle r}}_1)!\times {{\displaystyle r}}_1!}\right)\times \left(\frac{{{\displaystyle n}}_2!}{({{\displaystyle n}}_2-{{\displaystyle r}}_2)!\times {{\displaystyle r}}_2!}\right)$

Calculation:

Using the formula of permutation and combination-

  $\left(\frac{{{\displaystyle n}}_1!}{{{\displaystyle (n}}_1-{{\displaystyle r}}_1)!\times {{\displaystyle r}}_1!}\right)\times \left(\frac{{{\displaystyle n}}_2!}{({{\displaystyle n}}_2-{{\displaystyle r}}_2)!\times {{\displaystyle r}}_2!}\right)$

Now, putting the values in the above equation we get -

  $\begin{array}{l}=\left(\frac{8!}{(8-2)!\times 2!}\right)\times \left(\frac{2!}{(2-1)!\times 1!}\right)\\ =\frac{8\times 7\times 6!}{6!\times 2\times 1}\times \left(\frac{2\times 1!}{1!\times 1}\right)\\ =8\times 7\\ =56\end{array}$

So, there are 56 ways to select one competitor and one alternative of 8 students.