6.3 Counting Principles, Permutations, and Combinations

Counting methods allow us to find the total number of outcomes in a sample space without writing them all down. The SHSAT will test your understanding of the Fundamental Counting Principle, factorials, permutations, and combinations. The most critical skill in this section is deciding whether the order of selection matters.

The Fundamental Counting Principle

The Fundamental Counting Principle (FCP) states that if one event can occur in $m$ ways, a second event can occur in $n$ ways, and a third event can occur in $p$ ways, then the total number of ways the sequence of events can occur is the product: $\text{Total Outcomes} = m \times n \times p$

This principle applies when you are making a sequence of choices where each choice is selected from a separate, distinct group.

Example: A restaurant offers a lunch special with 3 appetizers, 5 main courses, and 4 desserts. How many different three-course meals can a customer create? Using the Fundamental Counting Principle, multiply the number of options for each course: $\text{Total Meals} = 3 \times 5 \times 4 = 60\text{ different meals}$

Factorials and Arranging Objects

A factorial (written as $n!$) represents the product of all positive integers from 1 up to $n$: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$ By definition, $0! = 1$.

Factorials are used when you want to arrange a complete set of $n$ distinct objects in a specific order. For instance, the number of ways to arrange 5 books on a shelf is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ways.

Permutations vs. Combinations: The Critical Distinction

When you are choosing a subset of objects from a larger group, you must ask yourself: Does the order of the selected items matter?

Permutations (Order Matters)

Use permutations when the arrangement, position, or specific role of the selected items makes a difference. The formula to select and arrange $r$ objects from a group of $n$ unique objects is: $P(n, r) = \frac{n!}{(n-r)!}$

Example: A club has 8 members. In how many ways can they elect a President, a Vice President, and a Treasurer?

1. Analyze the problem: Since the roles are distinct, order matters. This is a permutation of 8 items taken 3 at a time.
2. Apply the formula: $P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\text{ ways}$

Alternatively, think of this using the Fundamental Counting Principle: 8 choices for President, 7 remaining choices for Vice President, and 6 remaining choices for Treasurer. $8 \times 7 \times 6 = 336$.

Combinations (Order Does Not Matter)

Use combinations when you are forming a group, set, or committee where membership is all that matters, and the arrangement or order of selection is irrelevant. Because order does not matter, many permutations are considered duplicate groups. To eliminate these duplicates, divide the permutation formula by the number of ways to arrange the selected items ($r!$): $C(n, r) = \frac{n!}{r!(n-r)!}$

Example: A teacher needs to select 3 students from a class of 8 to help carry supplies. In how many ways can this group of students be chosen?

1. Analyze the problem: The group of students has no individual roles; a group of {Alice, Bob, Charlie} is identical to {Charlie, Bob, Alice}. Therefore, order does not matter. This is a combination of 8 items taken 3 at a time.
2. Apply the formula: $C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\text{ ways}$

Notice how $C(8,3)$ (56 ways) is much smaller than $P(8,3)$ (336 ways) because we divided by $3! = 6$ to remove the duplicates caused by ordering.

Choosing Multiple Groups (Compound Combinations)

Sometimes, an SHSAT question will ask you to select items from two separate categories to form a single group. In these cases, you calculate the combinations for each category separately and then multiply the results using the Fundamental Counting Principle.

For example, if a committee of 5 must be formed from a pool of 6 boys and 5 girls, and the committee must consist of exactly 3 boys and 2 girls:

1. Select the boys: The number of ways to choose 3 boys from 6 is $C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways.
2. Select the girls: The number of ways to choose 2 girls from 5 is $C(5, 2) = \frac{5 \times 4}{2 \times 1} = 10$ ways.
3. Find the total combinations: Multiply the two individual combinations: $20 \times 10 = 200$ different committees.

Breaking down complex counting problems into separate category selections is a highly effective strategy to avoid confusion.

Special Counting Scenarios

• The Handshake Problem: If there are $n$ people in a room and everyone shakes hands with everyone else exactly once, how many handshakes occur? Since a handshake involves 2 people and the order does not matter (A shaking hands with B is the same as B shaking hands with A), this is a combination of $n$ items taken 2 at a time: $C(n, 2) = \frac{n(n-1)}{2}$ For 10 people, the number of handshakes is $\frac{10 \times 9}{2} = 45$.
• Permutations with Repetition: If you are arranging letters in a word that has repeating letters, you must divide by the factorials of the frequencies of the repeating letters. The formula is: $\text{Arrangements} = \frac{N!}{n_1! \times n_2! \times \dots}$ For example, to arrange the letters of the word 'APPLES' (6 letters, with 'P' repeating 2 times): $\text{Arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360\text{ ways}$