10.2 Combinations, Permutations & Grouping

Order Versus Unordered Groups

Official Guide Math Review 3.4 asks you to count arrangements and selections. The entire skill is one question: does order matter?

If swapping two chosen people (or digits, or books) creates a different outcome — a ranking, a password, a labeled office — you are permuting. If swapping them creates the same committee, hand, or subset, you are combining.

On the 21-question, 45-minute, no-calculator section, write a short falling product (and divide by k! only when the group is unlabeled). You will not expand 12! in full.

Permutations: P(n, k)

P(n, k) = n! / (n − k)! = n × (n − 1) × ... × (n − k + 1)

You are lining up k distinct objects chosen from n distinct objects. The first slot has n options, the second n − 1, down to k slots.

Worked example. Eight candidates; choose a president, a vice president, and a treasurer, all different people.

P(8, 3) = 8 × 7 × 6 = 336.

Alice as president and Ben as VP is different from Ben as president and Alice as VP. Order matters, so do not divide by 3!. Dividing would produce C(8, 3) = 56, which counts unlabeled trios, not titled offices.

Worked example. How many 3-letter codes from 26 distinct letters, no repeats? P(26, 3) = 26 × 25 × 24 = 15,600. With repeats allowed it is not a permutation of distinct letters: 26³ = 17,576.

P(n, n) = n! / 0! = n!. Arranging all n objects is the factorial from 10.1. P(n, 1) = n. P(n, 0) = 1.

Cancel before you multiply: P(10, 4) = 10 × 9 × 8 × 7 = 5,040, which is 10! / 6!, not a request to write 3,628,800 / 720 by hand.

Combinations: C(n, k)

C(n, k) = n! / (k! (n − k)!) = P(n, k) / k!

The extra k! in the denominator erases order. Every unordered group of k people would have been counted k! times as a permutation — once for each way to line those k people up.

Worked example. A 4-person committee from 10 people.

C(10, 4) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 5,040 / 24 = 210.

If the stem had said "chair, secretary, treasurer, and member" as four distinct titles, you would have kept 5,040 = P(10, 4). Titles restore order.

Worked example. A 5-card hand from 12 distinct cards (a toy deck so the arithmetic stays small): C(12, 5) = (12 × 11 × 10 × 9 × 8) / 120 = 95,040 / 120 = 792.

Identities you actually use

• C(n, k) = C(n, n − k). Choosing k to include is choosing n − k to leave out. C(10, 8) = C(10, 2) = 45, which is faster than building C(10, 8) as an 8-factor product.
• C(n, 0) = C(n, n) = 1. One empty committee; one committee of everybody.
• C(n, 1) = n.
• C(n, 2) = n(n − 1) / 2. Pairs. Memorize this; it is the most common combination on Quant.

If k > n, C(n, k) = 0: you cannot choose 5 people from 4. That is an impossible case contributing 0, not a broken formula.

Groups of Groups

When a team has a stated makeup — 3 women and 2 men, 2 officers and 2 staff — multiply the combination counts for each subgroup. You are using FCP on the subgroups.

Worked example. 8 women and 6 men. A 5-person committee with exactly 3 women (hence exactly 2 men).

C(8, 3) × C(6, 2) = 56 × 15 = 840.

Not C(14, 5) = 2,002, which ignores the gender quota. Not C(8, 3) = 56 alone, which never picks the men.

Worked example. 5 men and 4 women; a 4-person team with 2 of each, and then the 4 elect a chair from among themselves.

First choose the people: C(5, 2) × C(4, 2) = 10 × 6 = 60. Then choose the chair: 4 options. Total: 60 × 4 = 240.

The chair step is a 1-slot permutation among the already-chosen 4. Mixing "choose the group, then assign a title inside it" is standard GMAT glue.

If the groups themselves are labeled (Team A and Team B, or a day shift and a night shift), two assignments of the same people to opposite labels are different. If the groups are unlabeled (just "two pairs"), divide by the number of ways to shuffle identical group labels. GMAT usually labels the groups, so you multiply and stop.

Worked example of unlabeled pairs, because it catches people. Six people split into 2 unlabeled teams of 3.

C(6, 3) = 20 chooses Team 1, the rest are Team 2 — but that double-counts every partition, because the other trio could have been chosen first. Unlabeled: 20 / 2 = 10.

If the teams are "Room 1" and "Room 2," keep 20.

Restrictions: Must Include / Cannot Include

Name the restricted people first, then choose the rest from whoever remains eligible.

• Must include person A in a k-person committee from n people: A is seated. Choose k − 1 more from the other n − 1: C(n − 1, k − 1).
• Cannot include person B: drop B. Choose k from n − 1: C(n − 1, k).
• Must include A and cannot include B: seat A, drop B, choose k − 1 from n − 2: C(n − 2, k − 1).
• Exactly r of a special subgroup of size s: C(s, r) × C(n − s, k − r), provided 0 ≤ k − r ≤ n − s. If the second index is impossible, that case is 0.

Worked example. 12 people, 3 of whom are officers. A 4-person committee must include exactly 2 of the 3 officers.

C(3, 2) × C(9, 2) = 3 × 36 = 108.

If the stem said at least 2 officers, add the exactly-2 case to the exactly-3 case: 108 + C(3, 3) × C(9, 1) = 108 + 1 × 9 = 117.

If the stem said at most 1 officer: C(3, 0) × C(9, 4) + C(3, 1) × C(9, 3) = 126 + 3 × 84 = 126 + 252 = 378.

Worked example, must include / cannot include together. A 4-person committee from 10 people must include Ana and cannot include Ben. Seat Ana, drop Ben, choose 3 more from the remaining 8: C(8, 3) = 56.

A short note on circular arrangements

If n distinct people sit around a round table and only relative order matters (rotating everyone one seat is the same seating), the count is (n − 1)!. Fix one person as a reference, then arrange the other n − 1: 6 people around a table → 5! = 120.

GMAT Quantitative Reasoning rarely needs this. If seats are numbered, or there is a distinguished head of the table, it is an ordinary line: n!. If the problem never mentions a circle, do not divide by n. One sentence of awareness is enough; do not build a circular-permutation toolkit, and do not flip a necklace (that extra ÷ 2 is not a GMAT move).

Traps on Quant

• Using C(n, k) for titled roles (president / VP).
• Using P(n, k) for an unlabeled committee.
• Forgetting to divide P(n, k) by k! when converting to C(n, k).
• Computing C(n, k) when a quota requires a product of two combinations.
• "At least two" computed as a single C(s, 2) term instead of summing feasible r.
• Circular division on a row of seats.
• Treating C(n, k) with k > n as undefined rather than 0 for that case.