Problem 4

A traveling politician has to visit four cities in the state, in a specific order, but there are nine possible cities to choose from. In this problem order matters, so this is a permutation solution. How many ways are there to travel among four of the nine cities?

P(9, 4) = \frac{9!}{(9 -4)!} = \frac{9!}{5!} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot \cancel{5!}}{\cancel{5!}} = 3024

Problem 5

How many ways are there to form a new local charity committee if the committee consists of three local business owners and four community volunteers, if there are seven business owners and nine community volunteers to choose from?

So, we need to select three people from seven business owners, and four people from nine community volunteers. Order doesn’t matter, so these are not permutation problems, they are combination problems.

By the Rule of Product, the total number of ways to form a committee is 3-combinations of a set with 7 elements, multiplied by 4-combinations of a set of 9 elements.

C( 7, 3 ) \cdot C( 9, 4 ) = {7\choose 3} \cdot {9\choose 4} = \frac{7!}{3!(7 - 3)!} \cdot \frac{9!}{4!(9 - 4)!} = 35 \cdot 126 = 4,410

Problem 6

How many different strings can be made by reordering the letters of the word HORRORS?

This is not a permutation problem since some of the letters are duplicates. The set of letters contains three Rs, two Os, one H, and one S. The three Rs can be placed among the seven positions in C( 7, 3 ) ways. The two Os can be placed among the remaining four positions in C( 4, 2 ). The H can be placed among the remaining two positions in C( 2, 1 ). Finally, the S can be placed in the last singular position in C( 1, 1 ) ways.

By applying the Rule of Product, we can multiply the ways to pick the four letters together to get the answer.

C( 7, 3 ) \cdot C( 4, 2 ) \cdot C( 2, 1 ) \cdot C( 1, 1 )

= {7\choose 3} \cdot {4\choose 2} \cdot {2\choose 1} \cdot {1\choose 1}

= \frac{7!}{3!(7 - 3)!} \cdot \frac{4!}{2!(4 - 2)!} \cdot \frac{2!}{1!(2 - 1)!} \cdot \frac{1!}{1!(1 - 1)!}

= 35 \cdot 6 \cdot 2 \cdot 1 = 420