Both $P(n,0)$ and $C(n,0)$ equal 1. There is exactly one way to choose zero items, which is the empty set. By definition, $0! = 1$.

What if the selection allows replacement?

The standard $P(n,k)$ and $C(n,k)$ formulas do not apply. For ordered selections with replacement, the total is $n^k$. For unordered selections with replacement, the formula is $C(n+k-1, k)$.

How do I handle indistinguishable items in permutations?

If the pool of $n$ items contains subsets of identical items, divide $n!$ by the factorial of the count of each identical item: $\frac{n!}{n_1! n_2! \dots n_r!}$.

How to count outcomes with permutations and combinations

Permutations and combinations calculate the total number of ways to select $k$ items from a pool of $n$ distinct items without replacement. Use permutations when the specific sequence or order of selection matters, and combinations when the order does not matter.

The setup

Identify the total number of available items, $n$ , and the number of items being selected, $k$ . Determine whether the order of selection affects the outcome. If assigning roles, making sequences, or ranking items, order matters (permutation). If forming a group or set where roles are identical, order does not matter (combination).

The steps

If order matters, use the permutation formula: $P(n,k) = \frac{n!}{(n-k)!}$ .
If order does not matter, use the combination formula: $C(n,k) = \frac{n!}{k!(n-k)!}$ .
Expand the factorials. Cancel the common factorial terms in the numerator and denominator to simplify the calculation.
Multiply the remaining integers to find the total number of outcomes.

Checking the result

Verify that $n \ge k$ and both are non-negative integers. The final result must be a positive integer. For any $k > 1$ , $P(n,k)$ must be strictly greater than $C(n,k)$ .

Common errors

A frequent mistake is using combinations when the problem implies order, such as selecting a president and vice president instead of a generic committee. Another common error is failing to cancel factorials properly, leading to arithmetic errors when multiplying unnecessarily large numbers.

Worked example

A generic committee of 3 members is to be chosen from a group of 8 people. Then, a president, vice president, and treasurer are to be chosen from the same group of 8 people. Calculate the number of possible outcomes for both scenarios.

Part 1: Unordered selection (committee). Identify variables: $n = 8$ , $k = 3$ . Order does not matter, so use the combination formula: $C(8,3) = \frac{8!}{3!(8-3)!}$ $C(8,3) = \frac{8!}{3!5!}$ Expand and cancel the $5!$ term: $C(8,3) = \frac{8 imes 7 imes 6}{3 imes 2 imes 1}$ $C(8,3) = 56$

Part 2: Ordered selection (officers). Identify variables: $n = 8$ , $k = 3$ . Order matters, so use the permutation formula: $P(8,3) = \frac{8!}{(8-3)!}$ $P(8,3) = \frac{8!}{5!}$ Expand and cancel the $5!$ term: $P(8,3) = 8 imes 7 imes 6$ $P(8,3) = 336$

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References: OpenStax Introductory Statistics, Chapter 3 · Khan Academy: Probability and combinatorics