Chapter 4.4, Problem 33E

Evaluate each expression.

a. ₅C₂

b. ₈C₃

c. ₇C₄

d. ₆C₂

e. ₆C₄

To determine

The value of ${}_5{C}_2$.

Answer to Problem 33E

The value of ${}_5{C}_2$ is 10.

Explanation of Solution

Calculation:

Combinations:

A combination is an arrangement of n objects in r ways where the order of arrangement is not considered.

${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Where, n is the total number of objects and r is number of ways in which n objects can be selected.

Substitute n as 5 and r as 2.

$\begin{array}{c}{}_5{C}_2=\frac{5!}{\left(5-2\right)!2!}\\ =\frac{5\times 4\times 3\times \cancel{2!}}{3!\cancel{2!}}\\ =\frac{60}{6}\\ =10\end{array}$

Thus, the value of ${}_5{C}_2$ is 10.

To determine

The value of ${}_8{C}_3$.

Answer to Problem 33E

The value of ${}_8{C}_3$ is 56.

Explanation of Solution

Calculation:

Substitute n as 8 and r as 3.

$\begin{array}{c}{}_8{C}_3=\frac{8!}{\left(8-3\right)!3!}\\ =\frac{8\times 7\times 6\times 5\times 4\times \cancel{3!}}{5!\cancel{3!}}\\ =\frac{6,720}{120}\\ =56\end{array}$

Thus, the value of ${}_8{C}_3$ is 56.

To determine

The value of ${}_7{C}_4$.

Answer to Problem 33E

The value of ${}_7{C}_4$ is 35.

Explanation of Solution

Calculation:

Substitute n as 7 and r as 4.

$\begin{array}{c}{}_7{C}_4=\frac{7!}{\left(7-4\right)!4!}\\ =\frac{7\times 6\times 5\times \cancel{4!}}{3!\cancel{\cancel{4!}}}\\ =\frac{210}{6}\\ =35\end{array}$

Thus, the value of ${}_7{C}_4$ is 35.

To determine

The value of ${}_6{C}_2$.

Answer to Problem 33E

The value of ${}_6{C}_2$ is 15.

Explanation of Solution

Calculation:

Substitute n as 6 and r as 2.

$\begin{array}{c}{}_6{C}_2=\frac{6!}{\left(6-2\right)!2!}\\ =\frac{6\times 5\times \cancel{4!}}{\cancel{\cancel{4!}}2!}\\ =\frac{30}{2}\\ =15\end{array}$

Thus, the value of ${}_6{C}_2$ is 15.

To determine

The value of ${}_6{C}_4$.

Answer to Problem 33E

The value of ${}_6{C}_4$ is 15.

Explanation of Solution

Calculation:

Substitute n as 6 and r as 4.

$\begin{array}{c}{}_6{C}_4=\frac{6!}{\left(6-4\right)!4!}\\ =\frac{6\times 5\times \cancel{4!}}{\cancel{\cancel{4!}}2!}\\ =\frac{30}{2}\\ =15\end{array}$

Thus, the value of ${}_6{C}_4$ is 15.