Chapter 13.1, Problem 48E

a.

To determine

To calculate: The number of different ways to arrange cars.

Answer to Problem 48E

The number of different ways to arrange cars are $42!$ .

Explanation of Solution

Given information:

There are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Formula used:

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

Calculation:

Consider the provided information that there are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Total number of cars are,

  $12+10+15+5=42$ .

In order to arrange these 42 cars, use permutation as selection matters.

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

Here n is 42.

Required number of ways are $P\left(42,42\right)$ .

  $P\left(42,42\right)=42!$

Thus, the number of different ways to arrange cars are $42!$ .

b.

To determine

To calculate: The number of different ways to arrange 30 cars to make the train.

Answer to Problem 48E

The number of different ways to arrange 30 cars to make the trainare $\frac{42!}{12!30!}$ .

Explanation of Solution

Given information:

There are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Formula used:

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

Calculation:

Consider the provided information that there are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Total number of cars are,

  $12+10+15+5=42$ .

In order to arrange these 30 cars, use combination as selection does not matter.

Required number of ways are $C\left(42,30\right)$

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

Here n is 42 and r is 30.

  $\begin{array}{c}C\left(42,30\right)=\frac{42!}{30!\left(42-30\right)!}\\ =\frac{42!}{30!12!}\end{array}$

Thus, the number of different ways to arrange 30 cars to make the train are $\frac{42!}{12!30!}$ .

c.

To determine

To calculate: The number of different ways to form a train with 3 livestock cars, 6 flatcars, 6 tanker cars and 5 boxcars.

Answer to Problem 48E

The number of different ways to form a train with 3 livestock cars, 6 flatcars, 6 tanker cars and 5 boxcarsare $5.8\times {10}^9$ .

Explanation of Solution

Given information:

There are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Formula used:

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

Calculation:

Consider the provided information that there are 12 flatcars, 10 tanker cars, 15 boxcars and 5 livestock cars.

Total number of cars are,

  $12+10+15+5=42$ .

In order to form a train with 3 livestock cars, 6 flatcars, 6 tanker cars and 5 boxcars, use combination as selection does not matter.

Out of 5, 3 livestock cars are to be selected.

Out of 12, 6 flat cars are to be selected.

Out of 10, 6 tanker cars are to be selected.

Out of 15, 5boxcars are to be selected.

Required number of ways are $C\left(5,3\right)\cdot C\left(12,6\right)\cdot C\left(10,6\right)\cdot C\left(15,5\right)$

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

  $\begin{array}{c}C\left(5,3\right)\cdot C\left(12,6\right)\cdot C\left(10,6\right)\cdot C\left(15,5\right)=\frac{5!}{3!\left(5-3\right)!}\cdot \frac{12!}{6!\left(12-6\right)!}\cdot \frac{10!}{6!\left(10-6\right)!}\cdot \frac{15!}{5!\left(15-5\right)!}\\ =\frac{5!}{3!2!}\cdot \frac{12!}{6!6!}\cdot \frac{10!}{6!4!}\cdot \frac{15!}{5!10!}\\ =5.8\times {10}^9\end{array}$

Thus, the number of different ways to form a train with 3 livestock cars, 6 flatcars, 6 tanker cars and 5 boxcars are $5.8\times {10}^9$ .