Chapter 10.5, Problem 74E

a.

To determine

To calculate:

Number of combinations of three marbles can be drawn from the bag.

Answer to Problem 74E

Number of combinations of three marbles can be drawn from the bag s is $1$ .

Explanation of Solution

Given information:

  

Calculation:

The expression for combination for $n$ objects taken $r$ at a time is given by,

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

Now substitute $n=r=3$ in equation $\left(i\right)$ and get the output for the combination of $P{}_r=\frac{n!}{\left(n-r\right)!}$ marbles drawn from the bag.

  $\begin{array}{l}C{}_r=\frac{n!}{\left(n-r\right)!.r!}\\ C{}_r=\frac{3!}{\left(3-3\right)!.3!}\\ C{}_r=\frac{3!}{0!.3!}\\ C{}_r=1\end{array}$

Thus, answer is $1$ .

b.

To determine

To calculate:

Number of permutations of three marbles can be drawn from the bag.

Answer to Problem 74E

Number of permutations of three marbles can be drawn from the bag is $6$ .

Explanation of Solution

Given information:

  

Calculation:

The expression for combination for $n$ objects taken $r$ at a time is given by,

  $P{}_r=\frac{n!}{\left(n-r\right)!.r!}\mathrm{...}\left(ii\right)$

Now substitute $n=r=3$ in equation $\left(ii\right)$ and get the output for the combination of $P{}_r=\frac{n!}{\left(n-r\right)!}$ marbles drawn from the bag.

  $\begin{array}{l}P{}_r=\frac{n!}{\left(n-r\right)!.r!}\\ P{}_r=\frac{3!}{\left(3-3\right)!.3!}\\ P{}_r=6\end{array}$

Thus, answer is $6$ .