Question

be sure to show your work for full credit.

problem	fcp, p, or c?	work or formula
2. how many different ways can six students line up?
3. how many 3 member committees can be formed from a club with 15 people?

Explanation:

Problem 1

Step1: Identify the counting principle

We use the Fundamental Counting Principle (FCP) here. The FCP states that if there are \(n_1\) ways to do the first task, \(n_2\) ways to do the second task, \(\dots\), \(n_k\) ways to do the \(k\)-th task, then the total number of ways to do all the tasks together is \(n_1\times n_2\times\cdots\times n_k\).
Here, the number of ways to choose a shirt is \(n_1 = 4\), pants is \(n_2=3\), belts is \(n_3 = 2\), and shoes is \(n_4=1\).

Step2: Apply the FCP

The total number of outfits is the product of the number of choices for each item. So we calculate \(4\times3\times2\times1\).
First, \(4\times3=12\), then \(12\times2 = 24\), and \(24\times1=24\).

Step1: Identify the permutation

We need to find the number of permutations of 6 students. The formula for permutations of \(n\) distinct objects is \(P(n,n)=\frac{n!}{(n - n)!}=n!\), where \(n!=n\times(n - 1)\times\cdots\times1\). Here \(n = 6\).

Step2: Calculate \(6!\)

\(6!=6\times5\times4\times3\times2\times1\). Calculating step by step: \(6\times5 = 30\), \(30\times4=120\), \(120\times3 = 360\), \(360\times2=720\), \(720\times1 = 720\).

Step1: Identify the combination

We need to form 3 - member committees from 15 people. The formula for combinations is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 15\) (total number of people) and \(r = 3\) (number of people in the committee).

Step2: Calculate \(C(15,3)\)

First, calculate \(n!=15! = 15\times14\times13\times12!\), \(r!=3!=3\times2\times1 = 6\), and \((n - r)!=(15 - 3)!=12!\).
So \(C(15,3)=\frac{15!}{3!(15 - 3)!}=\frac{15\times14\times13\times12!}{6\times12!}\). The \(12!\) terms cancel out. Then \(\frac{15\times14\times13}{6}\). Calculate \(15\times14 = 210\), \(210\times13=2730\), then \(\frac{2730}{6}=455\).

Answer:

For Problem 1: FCP; Work: \(4\times3\times2\times1 = 24\); Number of outfits: \(24\)

Problem 2