Fundamental Counting Principle • Permutations • Combinations
Fundamental Counting Principle
If you have to make several choices, and each choice is independent of the others, just multiply the number of options for each choice.
Example:
You have 3 shirts and 4 pairs of pants.
Number of outfits = $3 \times 4 = 12$.
Permutations (order matters)
A permutation counts the number of ways to arrange things when order matters and you cannot use the same item twice
You multiply, but each time the number of choices gets smaller by 1.
Example:
Choosing President, VP, and Treasurer from 10 people:
$10 \times 9 \times 8$.
Common situations:
Combinations (order does not matter)
A combination counts the number of ways to choose a group of items when order does not matter.
You still multiply the choices, but then you divide by the number of ways you could have arranged the selected items (because those different arrangements are considered the same group).
Example:
Choosing a committee of 3 people from a group of 10:
$\dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
Common situations:
Quick decision guide: If the order of selection matters (positions, rankings, codes) → Permutation. If you are only choosing a group and roles are identical → Combination. If you are making independent choices from different categories → Fundamental Counting Principle.
Multiply the number of independent choices.
A smoothie shop offers 4 types of fruit, 3 types of yogurt, and 2 types of sweetener. How many different smoothies can be made if a customer chooses exactly one fruit, one yogurt, and one sweetener?
Each choice is independent. Multiply the number of options in each category:
$$4 \times 3 \times 2 = 24$$
Answer: 24
A restaurant lunch special lets a customer choose either a salad or one of 3 soups, then a sandwich that has 5 bread options and 4 filling options. How many different lunch specials are possible?
First choice: salad or 3 soups → \(1 + 3 = 4\) options.
Then sandwich: \(5\) breads \(\times 4\) fillings \(= 20\).
Total: \(4 \times 20 = 80\).
Answer: 80
A website requires a 4-character password. The first character must be a letter (A–Z), the second character must be a digit (0–9), and the last two characters can each be either a letter or a digit. How many different passwords are possible?
Position 1 (letter): 26 choices
Position 2 (digit): 10 choices
Position 3 (letter or digit): \(26 + 10 = 36\) choices
Position 4 (letter or digit): 36 choices
$$26 \times 10 \times 36 \times 36 = 336{,}960$$
Answer: 336,960
Order matters.
You multiply, but each time the number of choices gets smaller by 1.
A club has 12 members. Three distinct positions (president, secretary, and treasurer) will be filled by three different members. How many different ways can the positions be assigned?
Order matters and the same person cannot hold two positions:
$$P(12,3) = 12 \times 11 \times 10 = 1{,}320$$
Answer: 1,320
How many different 4-letter arrangements can be formed from the letters in the word MATH if no letter is repeated?
There are 4 distinct letters and we use all of them with no repetition:
$$P(4,4) = 4! = 4 \times 3 \times 2 \times 1 = 24$$
Answer: 24
Five students (Alex, Blake, Casey, Dana, and Ellis) are lining up for a photo. How many different line-ups are possible if Alex must stand at one of the two ends?
Alex has 2 choices for position (left end or right end).
The remaining 4 students can be arranged in the other 4 spots in \(4!\) ways:
$$2 \times 4! = 2 \times 24 = 48$$
Answer: 48
Order does not matter. Use \(C(n,r) = \dbinom{n}{r}\).
TI-84 Plus CE: How to calculate Combinations
Example: To calculate \(\dbinom{9}{4}\), type 9 → MATH → PRB → 3: nCr → 4 → ENTER
A teacher needs to choose 3 students out of a class of 10 to help set up for a school event. How many different groups of 3 students are possible?
Order does not matter (the group is the same regardless of selection order):
$$\dbinom{10}{3} = \dfrac{10\times9\times8}{3\times2\times1} = 120$$
Answer: 120
A committee of 5 people must be formed from a group of 8 juniors and 6 seniors. The committee must contain exactly 3 juniors and 2 seniors. How many different committees are possible?
Choose the juniors and seniors separately, then multiply:
$$\dbinom{8}{3} \times \dbinom{6}{2} = 56 \times 15 = 840$$
Answer: 840
From a group of 9 students, a team of 4 will be selected for a project. Two particular students, Mia and Noah, refuse to work together. How many different teams of 4 can be formed that do not include both Mia and Noah?
Method: Total − Unwanted Cases
Step 1: Total teams of 4 from 9 students (no restrictions):
$$\dbinom{9}{4} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$
Step 2: Teams that include both Mia and Noah (fix them on the team, then choose 2 more from the remaining 7):
$$\dbinom{7}{2} = \dfrac{7 \times 6}{2 \times 1} = 21$$
Step 3: Teams that do not include both:
$$126 – 21 = 105$$
Answer: 105
A club has 11 members. A committee of 5 people will be chosen. Two members, Jordan and Taylor, must both be on the committee or neither of them can be on it. How many different committees are possible under this restriction?
Method: Split into two valid cases
Case 1: Both Jordan and Taylor are on the committee. Fix them, then choose 3 more from the remaining 9 members:
$$\dbinom{9}{3} = \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$
Case 2: Neither Jordan nor Taylor is on the committee. Choose all 5 members from the remaining 9:
$$\dbinom{9}{5} = \dfrac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126$$
Total:
$$84 + 126 = 210$$
Answer: 210
From a group of 10 students (6 girls and 4 boys), a team of 5 will be selected. The team must include at least 2 boys. How many different teams are possible?
Method: Add the valid cases
We need teams with 2, 3, or 4 boys (since there are only 4 boys total).
2 boys + 3 girls:
$$\dbinom{4}{2} \times \dbinom{6}{3} = 6 \times 20 = 120$$
3 boys + 2 girls:
$$\dbinom{4}{3} \times \dbinom{6}{2} = 4 \times 15 = 60$$
4 boys + 1 girl:
$$\dbinom{4}{4} \times \dbinom{6}{1} = 1 \times 6 = 6$$
Total:
$$120 + 60 + 6 = 186$$
Answer: 186
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