Counting Methods

Counting Without Listing

Many GRE problems ask "how many ways…" — how many ways to arrange, select, seat, or assign. Listing them is hopeless once the numbers grow, so the Data Analysis section rewards a small toolkit: the fundamental counting principle, permutations, and combinations. The entire game is deciding whether order matters.

The Fundamental Counting Principle

If one choice can be made in $m$ ways and an independent second choice in $n$ ways, then the two together can be made in $m \times n$ ways. This extends to any number of stages: multiply the number of options at each stage.

A café offers $4$ breads, $3$ fillings, and $2$ sauces. A sandwich uses one of each.

The number of distinct sandwiches is $4 \times 3 \times 2 = 24$. The principle underlies everything else here — permutations and combinations are just organized applications of it.

Factorials

The number of ways to arrange $n$ distinct objects in a row is $n$ factorial:

$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$

So $5! = 120$. By convention $0! = 1$. Five books on a shelf can be ordered in $5! = 120$ ways, because there are $5$ choices for the first slot, $4$ for the next, and so on.

Permutations — Order Matters

A permutation counts arrangements of $r$ objects chosen from $n$ distinct objects when order matters:

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

Awarding gold, silver, and bronze to $3$ of $8$ runners is a permutation, because finishing first is different from finishing third: $_8P_3 = \dfrac{8!}{5!} = 8 \times 7 \times 6 = 336$.

Combinations — Order Does Not Matter

A combination counts selections of $r$ objects from $n$ when order does not matter:

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

Choosing a $3$-person subcommittee from $8$ people is a combination, because a committee is the same group regardless of the order you name its members: $_8C_3 = \dfrac{8!}{3!\,5!} = 56$. Notice $_8P_3 = 336$ is exactly $3! = 6$ times $_8C_3 = 56$: permutations overcount each selection by the $r!$ ways to order it.

Arrangements with Restrictions

Restrictions are where the GRE earns its difficulty. Two dependable techniques:

Glue technique (things must be together). Treat the required-together items as a single block, arrange the blocks, then multiply by the internal arrangements of the block. Seating $5$ people in a row with a specific couple adjacent: glue the couple into one unit → $4!$ arrangements of the $4$ units, times $2!$ for the couple's internal order $= 24 \times 2 = 48$.

Complement / slot technique (things must be apart, or a slot is fixed). Count all arrangements and subtract the forbidden ones, or fill the constrained slot first. The number of $4$-digit codes with no repeated digit, first digit nonzero: $9$ choices for the first digit, then $9, 8, 7$ for the rest $= 9 \times 9 \times 8 \times 7 = 4536$.

Identical Items

When some objects are identical, divide out their interchangeable arrangements. The number of distinct arrangements of the letters in LEVEL ($5$ letters with $L$ twice and $E$ twice) is $\dfrac{5!}{2!\,2!} = 30$.

A Reliable Game Plan

• Does order matter? Yes → permutation. No → combination.
• One choice per category? Multiply the category sizes.
• Must-be-together? Glue into a block, then multiply by the block's internal order.
• Must-be-apart or a fixed slot? Use the complement, or fill the constrained slot first.
• Repeated identical items? Divide by the factorial of each repeat count.