Chapter 13.1, Problem 28E

To determine

To calculate: The value of expression $\frac{P\left(6,3\right)}{P\left(4,2\right)}$ .

Answer to Problem 28E

The value of expression $\frac{P\left(6,3\right)}{P\left(4,2\right)}$ is $10$ .

Explanation of Solution

Given information:

The expression $\frac{P\left(6,3\right)}{P\left(4,2\right)}$ .

Formula used:

If there are n objects taken r at a time then permutation is defined as $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$ .

Calculation:

Consider the expression $\frac{P\left(6,3\right)}{P\left(4,2\right)}$ .

First evaluate the numerator that is $P\left(6,3\right)$ .

Recall that if there are n objects taken r at a time then permutation is defined as $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$ .

Here n is 6 and r is 3.

  $\begin{array}{c}P\left(6,3\right)=\frac{6!}{\left(6-3\right)!}\\ =\frac{6!}{3!}\\ =\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}\\ =120\end{array}$

Next evaluate the denominator that is $P\left(4,2\right)$ .

Recall that if there are n objects taken r at a time then permutation is defined as $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$ .

Here n is 4 and r is 2.

  $\begin{array}{c}P\left(4,2\right)=\frac{4!}{\left(4-2\right)!}\\ =\frac{4!}{2!}\\ =\frac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1}\\ =12\end{array}$

Therefore,

  $\begin{array}{c}\frac{P\left(6,3\right)}{P\left(4,2\right)}=\frac{120}{12}\\ =10\end{array}$

Thus, the value of expression $\frac{P\left(6,3\right)}{P\left(4,2\right)}$ is $10$ .