Chapter 4.CR, Problem 9CR

To determine

To find:

The number of room assignments can Mrs. Viant possibly have for the rooms 100, 101, 102, and 103.

Answer to Problem 9CR

Solution:

The number of different room assignment is 8870400.

Explanation of Solution

Given information:

In planning the room assignments for the overnight field trip, Mrs. Viant needs to put 3 girls and 1 chaperone in each room. There are 12 girls and 4 chaperones on the trip. There are four rooms 100, 101, 102, and 103.

Approach:

For selection of objects from a group without regard to their arrangement is the combination.

Formula for combination is used for counting number of selection of objects from a group without regard to their arrangement is the combination.

Formula used:

Formula to calculate the possible number of count to select $r$ objects from $n$ objects in combination is,

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

$n!=n\cdot \left(n-1\right)\cdot \left(n-2\right)\mathrm{...3}\cdot 2\cdot 1$

Here $n$ and $r$ both are positive integers.

Calculation:

Here choice of chaperones and girls is independent of order.

Use formula of combination to get total number of selection.

Total number of girls is 12 and total number of chaperones is 4.

Mrs. Viant needs to put 3 girls and 1 chaperone in each room.

Use the formula for combination to calculate the possible number of ways.

For room 100 number of combination of girls is,

Substitute 12 for $n$ and 3 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_{12}{C}_3& =& \frac{12!}{3!\left(12-3\right)!}\\ & =& \frac{12!}{3!9!}\\ & =& \frac{12\times 11\times 10\times 9!}{3\times 2\times 1\times 9!}\\ & =& 220\end{array}$

There are 220 different ways to choose 3 girls out of 12.

For room 100 number of combination of chaperones is,

Substitute 4 for $n$ and 1 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_4{C}_1& =& \frac{4!}{1!\left(4-1\right)!}\\ & =& \frac{4!}{1!3!}\\ & =& \frac{4\times 3!}{1\times 3!}\\ & =& 4\end{array}$

There are 4 different ways to choose 1 chaperone out of 4.

Total number of ways to choose 3 girls and 1 chaperone for room 100 is,

$\begin{array}{r}220\times 4\\ =880\end{array}$

There are 880 numbers of ways to select a group for room no 100.

There are more 9 girls and 3 chaperones left.

For room 101 number of combination of girls is,

Substitute 9 for $n$ and 3 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_9{C}_3& =& \frac{9!}{3!\left(9-3\right)!}\\ & =& \frac{9!}{3!6!}\\ & =& \frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\\ & =& 84\end{array}$

There are 84 different ways to choose 3 girls out of 9.

For room 101 number of combination of chaperones is,

Substitute 3 for $n$ and 1 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_3{C}_1& =& \frac{3!}{1!\left(3-1\right)!}\\ & =& \frac{3!}{1!2!}\\ & =& \frac{3\times 2!}{1\times 2!}\\ & =& 3\end{array}$

There are 3 different ways to choose 1 chaperone out of 3.

Total number of ways to choose 3 girls and 1 chaperone for room 101 is,

$\begin{array}{l}84\times 3\\ =252\end{array}$

There are 252 numbers of ways to select a group for room no 101.

There are more 6 girls and 2 chaperones left.

For room 102 number of combination of girls is,

Substitute 6 for $n$ and 3 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_6{C}_3& =& \frac{6!}{3!\left(6-3\right)!}\\ & =& \frac{6!}{3!3!}\\ & =& \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\\ & =& 20\end{array}$

There are 20 different ways to choose 3 girls out of 6.

For room 102 number of combination of chaperones is,

Substitute 2 for $n$ and 1 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_2{C}_1& =& \frac{2!}{1!\left(2-1\right)!}\\ & =& \frac{2!}{1!1!}\\ & =& \frac{2\times 1!}{1\times 1!}\\ & =& 2\end{array}$

There are 2 different ways to choose 1 chaperone out of 2.

Total number of ways to choose 3 girls and 1 chaperone for room 102 is,

$\begin{array}{l}20\times 2\\ =40\end{array}$

There are 40 numbers of ways to select a group for room no 102.

There are more 3 girls and 1 chaperones left.

For room 103 number of combination of girls is,

Substitute 3 for $n$ and 3 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_3{C}_3& =& \frac{3!}{3!\left(3-3\right)!}\\ & =& \frac{3!}{3!0!}\\ & =& 1\end{array}$

There is 1 way to choose 3 girls out of 3.

For room 103 number of combination of chaperones is,

Substitute 1 for $n$ and 1 for $r$ in the formula of combination.

Here $n$ and $r$ both are positive integers.

$\begin{array}{rll}{}_1{C}_1& =& \frac{1!}{1!\left(1-1\right)!}\\ & =& \frac{1!}{1!0!}\\ & =& 1\end{array}$

There is 1 different way to choose 1 chaperone out of 1.

Total number of ways to choose 3 girls and 1 chaperone for room 103 is,

$\begin{array}{l}1\times 1\\ =1\end{array}$

There is 1 way to select a group for room no 103.

The total number of ways for the assignment is the product of each selection.

Total number of ways is,

$\begin{array}{l}880\times 252\times 40\times 1\\ =8870400\end{array}$

Conclusion:

Thus, the number of different room assignment is 8870400.