Combinatorics and probability

This chapter covers the following:

• Permutations
• Combinations
• Key differences
• Special cases
• Probability
• Law of total probability
• Law of compound or joint probability
• Bayes’ theorem

In mathematics, permutations and combinations are counting methods. You use them to figure out how many ways you can arrange or select items. These formulas also appear in the FE Reference Handbook’s Mathematics section, under “Permutations and Combinations,” so you can look them up rather than memorize them.

Permutations

Permutations count the number of ways to arrange objects when order matters.

Formula for permutations (without repetition)

Use this when you’re arranging $r$ items chosen from $n$ distinct items, with no repeats:

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

Where:

• $n!$ (“$n$ factorial”) means $n\times (n-1)\times \cdots \times 1$
• $r$ is the number of items you arrange

Example

How many ways can you arrange 3 letters from the word “MATH”?

• Number of distinct letters: $n=4$ (M, A, T, H)
• Number of letters to arrange: $r=3$

$$P(4,3)=\frac{4!}{(4-3)!}=\frac{24}{1}=24$$

So, there are 24 ways to arrange 3 letters out of 4.

Combinations

Combinations count the number of ways to select objects when order doesn’t matter.

Formula for combinations

Use this when you’re choosing $r$ items from $n$ distinct items and the order of the chosen items is irrelevant:

$$C(n,r)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$$

Example

How many ways can you choose 3 students from a group of 5?

• $n=5$
• $r=3$

$$C(5,3)=\frac{5!}{3!(5-3)!}=\frac{120}{6\times 2}=\frac{120}{12}=10$$

So, there are 10 ways to choose 3 students from 5.

Key differences

Aspect	Permutations	Combinations
Order matters?	Yes	No
Formula	$\frac{n!}{(n-r)!}$	$\frac{n!}{r!(n-r)!}$
Example	Arranging medals	Choosing committee members

Watch out:

• Order matters or not? Swapping two chosen items into a different outcome (like gold vs. silver) → use a permutation. Swapping them changes nothing (like committee members) → use a combination.
• Mutually exclusive vs. independent. Mutually exclusive means the two events can’t both happen - that’s what lets the addition rule below drop its overlap term. Independent means one event doesn’t change the other’s probability ($P(A\mid B)=P(A)$) - a different idea, used in the joint probability rule.

Special cases

Permutations with repetition

Use this when each of the $r$ positions can be filled by any of the $n$ items (so repeats are allowed):

$$n^r$$

Example: 4-digit PIN code using digits 0-9:

• $n=10$ possible digits
• $r=4$ positions

$$10^4=10,000\text{ possible codes}$$

Permutations with identical items

If some objects are indistinguishable, swapping those identical objects doesn’t create a new arrangement. That reduces the number of distinct permutations.

The number of different permutations of $n$ objects, where:

• $n_1$ are of type 1
• $n_2$ are of type 2
• …
• $n_k$ are of type $k$
• and ${\sum }_{i=1}^k{n}_i=n$

Then the number of distinct permutations is:

$$\frac{n!}{n_1!\cdot n_2!\cdots n_k!}$$

Example

How many distinct ways can you arrange the letters in the word BALLOON?

Letters: B, A, L, L, O, O, N

• $n=7$
• L appears 2 times
• O appears 2 times

So,

$$\frac{7!}{2!\cdot 2!}=\frac{5040}{4}=1260$$

Probability

Definition: Probability measures how likely an event is to occur.

Let $S$ be the sample space (all possible outcomes) and $A$ an event (a set of outcomes). The probability of event $A$, written $P(A)$, is:

$$P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

Example: If a fair die is rolled, what is the probability of getting a 4?

$$P(4)=\frac{1}{6}$$

A combination can also serve as the favorable-outcome count in a probability problem:

Example: exactly 3 heads in 4 flips

If you flip a fair coin 4 times, what’s the probability of getting exactly 3 heads?

• Total outcomes: $2^4=16$
• Favorable outcomes: choose which 3 of the 4 flips are heads, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$

$$P(\text{3 heads})=\frac{\left(\genfrac{}{}{0ex}{}{4}{3}\right)}{2^4}=\frac{4}{16}=\frac{1}{4}$$

Answer: $\frac{1}{4}$, or 25%.

Addition rule (union of events)

Use this for the probability that at least one of two events happens:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

Subtracting $P(A\cap B)$ avoids double-counting outcomes shared by both events. When $A$ and $B$ are mutually exclusive (no shared outcomes), $P(A\cap B)=0$ and this reduces to:

$$P(A\cup B)=P(A)+P(B)$$

Example: heart or face card

A standard deck has 52 cards. What’s the probability of drawing a card that is a heart or a face card (jack, queen, king)?

• $P(\text{heart})=\frac{13}{52}$
• $P(\text{face card})=\frac{12}{52}$
• $P(\text{heart and face card})=\frac{3}{52}$ (the jack, queen, and king of hearts)

$$P(\text{heart}\cup \text{face card})=\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}\approx 0.423$$

Answer: approximately $0.423$, or about 42.3%.

Law of total probability

If events $B_1,B_2,...,B_n$ are mutually exclusive and exhaustive (meaning they partition the sample space), then any event $A$ can be found by adding up the ways $A$ can happen through each case $B_i$:

$$P(A)=\sum _{i=1}^{n}P(A\mid B_i)P(B_i)$$

Example: total probability of a defective item

Suppose a factory has two machines:

• Machine 1 produces 60% of the items and has a defect rate of 1%.
• Machine 2 produces 40% of the items and has a defect rate of 2%.

What is the probability that a randomly selected item is defective?

Let:

• $B_1$: item from Machine 1, $P(B_1)=0.6$
• $B_2$: item from Machine 2, $P(B_2)=0.4$
• $A$: item is defective

Then,

$$\begin{array}{rl}P(A)& =P(A\mid B_1)P(B_1)+P(A\mid B_2)P(B_2)\\ & =(0.01)(0.6)+(0.02)(0.4)=0.006+0.008=0.014\end{array}$$

Answer: the probability of a defective item is $0.014$, or $1.4\%$.

Law of compound or joint probability

The joint probability of two events $A$ and $B$ occurring together is:

$$P(A\cap B)=P(A\mid B)\cdot P(B)=P(B\mid A)\cdot P(A)$$

This formula connects:

• a conditional probability (like $P(A\mid B)$)
• the probability of the condition (like $P(B)$)
• the probability of both events happening (like $P(A\cap B)$)

For example, if $P(B)=0.5$ and $P(A\mid B)=0.3$, then $P(A\cap B)=0.3\times 0.5=0.15$. You’ll see this same multiplication at work in the Bayes’ theorem example below, where it builds the numerator of the reversed conditional probability.

Bayes’ theorem

Bayes’ theorem lets you reverse a conditional probability. In other words, it helps you find $P(A\mid B)$ when you know $P(B\mid A)$.

Given events $A$ and $B$ with $P(B)>0$:

$$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$$

Using the law of total probability in the denominator:

$$P(A_i\mid B)=\frac{P(B\mid A_i)\cdot P(A_i)}{{\sum }_{j=1}^{n}P(B\mid A_j)\cdot P(A_j)}$$

Exam tip: Carry full precision through every intermediate step of a multi-step probability problem, and round only the final answer. FE distractors are often spaced closely enough that rounding early (like truncating $P(A)$ before dividing) can land you on the wrong choice. You’ll find the addition rule and Bayes’ theorem listed under “Engineering probability and statistics” in the FE Reference Handbook - look them up there rather than memorize them, just like the counting formulas above.

Example: which machine produced the defective item?

This continues the factory example above. What is the probability that a defective item came from Machine 2?

We already know:

• $P(A)=0.014$
• $P(B_2)=0.4$
• $P(A\mid B_2)=0.02$

Now apply Bayes’ theorem:

$$P(B_2\mid A)=\frac{P(A\mid B_2)\cdot P(B_2)}{P(A)}=\frac{0.02\cdot 0.4}{0.014}=\frac{0.008}{0.014}\approx 0.5714$$

Answer: about $57.14\%$ of defective items come from Machine 2.