Counting problems, permutations, and combinations

This section connects closely to probability. Many counting problems can be solved using the same step-by-step multiplication you used in probability. Permutations, however, often use a memorized formula.

Combinations

Probabilities tell you how likely an event is. Combinations (in this section’s sense) tell you how many possible outcomes there are.

In a probability problem, you often multiply fractions. The result gets smaller as the event becomes more specific.

Counting combinations works in a similar way, but you multiply whole numbers instead of fractions. For example:

• If there are 2 equally likely possibilities, the probability of one specific outcome is $1/2$.
• The corresponding count of possibilities is $2$.

Let’s start with an easy example:

There are $6$ cars and $4$ drivers. How many possible combinations are there of drivers in their respective cars?

If this were a probability question, you might say:

• Probability of picking a particular car: $1/6$
• Probability of picking a particular driver: $1/4$

For counting combinations, you use the reciprocals as counts:

• $6$ choices for the car
• $4$ choices for the driver

Multiply to get the total number of combinations:

• $6\times 4=24$

Permutations

Permutations are more complex than combinations because order matters.

• For combinations, the order of the items does not matter.
• For permutations, the order of the items in the arrangement does matter.

The formula to determine a permutation ($P$) is:

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

where $n$ is the number of objects in the problem and $r$ is the number of things we take from the total.

This formula introduces a new symbol, the factorial ($!$). A factorial means you multiply the number by every whole number below it down to $1$. For example:

• $5!=5\ast 4\ast 3\ast 2\ast 1=120$

You can compute factorials by writing them out, or by using your calculator. On a TI-84, the factorial symbol is found under the “math” button and the “probability” tab.

Here’s an example:

There is a table with four seats and five possible guests to fill it. How many possible combinations of seat arrangements are there?

We’ll first use the formula, then connect it to the counting idea.

Remember the formula: $P=\frac{n!}{(n-r)!}$

For this question, $n=5$ and $r=4$.

$$\begin{array}{rl}P& =\frac{5!}{(5-4)!}\\ & \\ P& =120\end{array}$$

Now, here’s what that means in plain counting terms:

• The first seat has $5$ choices.
• After one person sits, the second seat has $4$ choices.
• The third seat has $3$ choices.
• The fourth seat has $2$ choices.

So the total number of arrangements is:

• $5\times 4\times 3\times 2=120$

That’s why permutations count “combinations that depend on order.”