Permutations, Combinations, and Factorials: Advanced Counting in Probability 3.4

Section 3.4 Additional Topics in Probability and Counting

Factorials

Factorials are fundamental in counting problems, especially when determining the number of ways to arrange or select objects. The factorial of a non-negative integer n, denoted n!, is the product of all positive integers less than or equal to n.

• Definition: For any integer n ≥ 0, n! is defined as the product of all positive integers up to n.

• Special Cases: 0! = 1 and 1! = 1 by definition.

• Example: 5! = 5 × 4 × 3 × 2 × 1 = 120

Calculator Tip: Use the PRB (probability) button to access the factorial function (!).

Permutations (nPr)

Permutations count the number of ways to arrange r objects selected from n distinct objects, where order matters.

• Definition: An ordered arrangement of objects.

• Formula:

$nPr = \frac{n!}{(n - r)!}$

• Example: The number of ways to award 1st, 2nd, and 3rd place among 20 contestants: $20P3 = \frac{20!}{(20-3)!} = 6840$ ways.

Combinations (nCr)

Combinations count the number of ways to select r objects from n distinct objects, where order does not matter.

• Definition: An unordered selection of objects.

• Formula:

$nCr = \frac{n!}{(n - r)! \ r!}$

• Example: The number of ways to choose 6 numbers from 52 in a lottery: $52C6 = \frac{52!}{(52-6)!6!} = 20,358,520$ ways.

Distinguishable Permutations

When some objects are identical, the number of distinguishable arrangements is less than the total number of permutations. This is important for problems involving repeated items.

• Definition: The number of ways to arrange n objects, where there are groups of identical items.

• Formula:

$\text{Number of distinguishable permutations} = \frac{n!}{n_1! \times n_2! \times n_3! \times \cdots \times n_k!}$

• Where $n_1, n_2, ..., n_k$ are the counts of each type of repeated object, and $n_1 + n_2 + ... + n_k = n$.

• Example: Arranging the letters in "AABBB": $\frac{5!}{2! \times 3!} = 10$ ways.

Summary Table: When to Use Each Counting Method

Situation	Key Features	Method	Formula
Arranging all items (no repeats)	One number given, or r = n	Factorial	$n!$
Selecting r items from n, order matters	Two numbers, order matters	Permutation	$\frac{n!}{(n - r)!}$
Selecting r items from n, order does not matter	Two numbers, order does not matter	Combination	$\frac{n!}{(n - r)! r!}$
Arranging n items with repeats	Repeated items	Distinguishable Permutation	$\frac{n!}{n_1! n_2! ... n_k!}$

Counting and Probability

Counting methods are often used to determine the number of possible outcomes in a sample space, which is essential for calculating probabilities.

• Sample Space: The set of all possible outcomes.

• Probability of an Event E:

$P(E) = \frac{\text{Number of outcomes in event E}}{\text{Number of outcomes in sample space}}$

• Example: Probability of guessing a unique 4-digit code (no repeats): $P(E) = \frac{1}{4!} = \frac{1}{24} \approx 0.04167$

Worked Examples

• Committees: Number of ways to choose 4 students from 15 (order not important): $15C4 = 1365$

• Race Results: Number of ways 20 people can finish 1st, 2nd, 3rd (order matters): $20P3 = 6840$

• Arranging Disks with Repeats: 4 red, 6 black, 2 green, 3 yellow disks in a row: $\frac{15!}{4!6!2!3!} = 6,306,300$

• All Arrangements: 8 skiers, all finish orders: $8! = 40,320$

• Lottery: Ways to choose 6 numbers from 52 (order not important): $52C6 = 20,358,520$

• Probability of Winning Lottery: $P(\text{win}) = \frac{1}{20,358,520} \approx 0.000000049$

• Singing Competition: 16 finalists, top 5 prizes (order matters): $16P5 = 524,160$

• Necklace with Repeated Glass: 9 blue, 3 red, 7 green pieces: $\frac{19!}{9!3!7!} = 11,085,360$

• Pizza Toppings Probability: Probability of a specific 3-topping pizza from 9 toppings: $9C3 = 84$ possible pizzas, $P = \frac{1}{84} \approx 0.012$

• Cyclists Race: 15 cyclists, top 3 finishers: $15P3 = 2730$

• Basketball Defense Assignments: 5 players, each defends one: $5! = 120$ or $5P5 = 120$

• Restaurant Meal Combinations: 3 appetizers (10 choices), 2 main courses (18), 2 desserts (13): $10C3 \times 18C2 \times 13C2 = 120 \times 153 \times 78 = 1,432,080$

• Soil Sample Selection: Choose 2 contaminated from 8, 8 non-contaminated from 42: $8C2 \times 42C8 = 28 \times 118,030,185 = 3,304,845,180$

Key Takeaways

• Use factorials for arranging all items or when r = n.

• Use permutations when order matters and selecting r from n.

• Use combinations when order does not matter and selecting r from n.

• Use distinguishable permutations when arranging items with repeats.

• Counting methods are essential for calculating probabilities in finite sample spaces.

Additional info: The notes also emphasize calculator usage for these operations, especially for large numbers, and the importance of entering expressions in one step to avoid overflow or errors.