Chapter 4.4, Problem 69EC

Seating in a Movie Theater How many different ways can 5 people—A, B, C, D, and E—sit in a row at a movie theater if (a) A and B must sit together; (b) C must sit to the right of, but not necessarily next to, B; (c) D and E will not sit next to each other?

To determine

The numberof ways where A and B can sit together.

Answer to Problem 69EC

The number of ways where A and B can sit together is 48 ways.

Explanation of Solution

Given info:

There are 5 members A, B, C, D and E and these members should sit in a row.

Calculation:

The formula used to find the factorial of any number n is given below:

$\begin{array}{l}n!=n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(1\right)\\ 0!=1\end{array}$

The number of ways where A and B can sit together is calculated as follows:

There will be 3 single seats and one seat will be occupied by A and B.

$\left(\text{AB}\right),\left(\text{C}\right),\left(\text{D}\right),\left(\text{E}\right)$

Substitute n as 4,

$\begin{array}{c}4!=4\left(4-1\right)\left(4-2\right)\mathrm{...}\left(1\right)\\ =24\end{array}$

Also, there is another possibility where a single seat is occupied by B and A. This will also have the same number of arrangements.

Thus, the number of ways where A and B can sit together is $24+24=48\text{ Ways}$.

To determine

The number of ways where C has to sit right next to anyone but not necessarily next to B.

Answer to Problem 69EC

There are 60 ways to make C to sit right next to anyonebut not necessarily next to B.

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.

The number of ways where C has to sit right next to anyone but not necessarily next to B is given below:

Substitute n as 5 and r as 2.

$\begin{array}{c}{}_5{C}_2=\frac{5!}{\left(5-2\right)!2!}\\ =\frac{5\times 4\times \cancel{3!}}{\cancel{3!}2!}\\ =\frac{20}{2}\\ =10\end{array}$

Also, the remaining three places can be occupied in 3! Ways, which is 6 ways.

Fundamental counting rule:

The number of ways in which a sequence of n events occur if the first event can occur in $k_1$ ways, the second event can occur in $k_2$ ways and so on is given by $k_1\times k_2\times \mathrm{...}\times k_n$

The number of ways where C has to sit right next to anyone but not necessarily next to B is

$\begin{array}{c}{}_5{C}_2\times 3!=10\times 6\\ =60\end{array}$

Thus, there are 60 ways to make C to sit right next to anyone.

To determine

The number of ways where D and Ewill not sit next to each other.

Answer to Problem 69EC

The number of ways where D and E will not sit next to each other is 72 ways.

Explanation of Solution

Calculation:

The formula used to find the factorial of any number n is given below:

$\begin{array}{l}n!=n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(1\right)\\ 0!=1\end{array}$

The total number of ways to arrange 5 members in a row will be in 5! ways which 120 ways.

The number of ways where D and E can sit together is calculated as follows:

There will be 3 single seats and two seats will be occupied by D and E.

$\left(\text{DE}\right),\left(\text{A}\right),\left(\text{B}\right),\left(\text{C}\right)$

Substitute n as 4,

$\begin{array}{c}4!=4\left(4-1\right)\left(4-2\right)\mathrm{...}\left(1\right)\\ =24\end{array}$

Also, there is another possibility where a single seat is occupied by E and D. This will also have the same number of arrangements.

The number of ways where D and E can sit together is $24+24=48\text{ Ways}$.

Finally, the number of ways where D and E will not sit together is given below:

$\begin{array}{c}\left(\begin{array}{l}\text{Number of ways where}\\ \text{D and E will not sit }\\ \text{together}\end{array}\right)=\left(\begin{array}{l}\text{Total number of ways}\\ \text{to arrange 5 members}\end{array}\right)-\left(\begin{array}{l}\text{Total number of ways}\\ \text{where D and E can sit together}\end{array}\right)\\ =120-48\\ =72\end{array}$

Thus, there are 72 ways where D and E will not sit together.