Combinations and permutations: complete GMAT guide

GMAT counting questions ask for the number of ways to arrange or select objects. They appear in Problem Solving and occasionally in Data Sufficiency. The content is built from four tools:

• The fundamental counting principle
• Permutations, where order matters
• Combinations, where order does not matter
• Case splits and complementary counting

You do not need advanced combinatorics. You need to read the restrictions carefully and avoid counting the same outcome twice.

If one decision can be made in $m$ ways and a second independent decision can be made in $n$ ways, the two decisions together can be made in $mn$ ways. Extend the rule to as many decisions as the problem has.

The rule applies only when the decisions are independent. If choosing one item removes it from the pool, the number of choices for the next decision changes.

A permutation counts arrangements. If you choose $r$ objects from $n$ distinct objects and the order matters, the number of arrangements is

For example, the number of ways to choose a president, vice president, and treasurer from 10 people is $10 \times 9 \times 8=720$. The same three people in different offices count as different outcomes.

When all $n$ objects are arranged, the number of orders is $n!$. The notation $n!$ means the product of all positive integers from 1 to $n$. By definition, $0!=1$.

Arrangements with restrictions

Handle restrictions by placing the restricted objects first. If a president and vice president must be chosen from 6 men and 4 women, and the president must be a woman, choose the president in 4 ways, then the vice president in 9 ways. The answer is 36, not $10 \times 9=90$.

If two people must sit together, treat them as one block. Arrange the block with the other objects, then multiply by the number of orders inside the block. If two people must not sit together, count all arrangements and subtract the arrangements where they do sit together.

A combination counts selections. If you choose $r$ objects from $n$ distinct objects and the order does not matter, the number of selections is

For example, the number of ways to choose a committee of 3 from 8 people is $\frac{8!}{3!5!}=56$. Choosing Alice, Ben, and Carla is the same committee as choosing Carla, Ben, and Alice.

Two symmetry rules save time:

• $\binom{n}{r}=\binom{n}{n-r}$. Choosing 3 people to include is the same as choosing $n-3$ people to exclude.
• $\binom{n}{0}=\binom{n}{n}=1$. There is one way to choose none or all.

The phrase "at least one" usually signals complementary counting. Count the opposite case and subtract from the total. This is faster than adding the cases with exactly one, exactly two, and exactly three women.

When some objects are identical, divide the total arrangements by the factorial of each repeated group. The number of distinct arrangements of a word with repeated letters is

where $a,b,c,\ldots$ are the counts of each repeated letter. For the word BANANA, there are 6 letters with 3 A's and 2 N's, so the number of distinct arrangements is $\frac{6!}{3!2!}=60$.

For $n$ distinct objects arranged in a circle, the number of arrangements is $(n-1)!$. The reason is that rotations of the same circle are not different arrangements, so you fix one object as a reference and arrange the remaining $n-1$ objects.

If clockwise and counterclockwise arrangements are considered the same, divide by 2. That situation is rare on the GMAT, but it appears in questions about necklaces or seating arrangements where the direction is explicitly ignored.

A grid path question asks how many shortest routes exist from one corner to another. If a route requires $a$ steps in one direction and $b$ steps in the other, every shortest route is a sequence of those steps, and the number of routes is $\binom{a+b}{a}$. For a grid that requires 3 steps right and 2 steps up, the number of shortest paths is $\binom{5}{3}=10$. The same logic applies to sequences of wins and losses, or any situation where you arrange two types of identical moves.

Some restrictions are easier to handle by splitting into cases. If a committee must contain more women than men, count the valid compositions separately: 2 women and 1 man, or 3 women. Add the cases. Keep the cases mutually exclusive so you do not count the same committee twice.

Counting is often the first half of a probability question. If all outcomes are equally likely, probability is the number of favorable outcomes divided by the total number of outcomes. For example, the probability of choosing 2 women from 4 women and 3 men when selecting 2 people is

Make sure the numerator and denominator count the same kind of outcome. If the denominator counts committees, the numerator must count committees, not ordered pairs.

1. Decide whether order matters. If rearranging the same items creates a different outcome, use permutations. If not, use combinations.
2. Write the pool size and the number chosen. Check whether items can be reused.
3. Handle restrictions by placing restricted items first, grouping items that must stay together, or subtracting forbidden cases.
4. For "at least" questions, count the complement.
5. For repeated items, divide by the factorials of the repeated groups.
6. Check for double counting before you add cases.

• Using combinations when order matters, or permutations when it does not.
• Counting the same committee in different orders.
• Forgetting that choosing 3 people to include is the same as choosing the rest to exclude.
• Adding cases that overlap.
• Treating identical objects as distinct.
• Forgetting to multiply by the internal arrangements of a block.
• Using $n!$ for a circular arrangement instead of $(n-1)!$.