Combinations
1. Core Idea
A combination is a selection where order does not matter. Choosing Amit, Bina and Chetan for a committee is the same committee no matter what sequence you name them in.
The formula comes straight from permutations: count the ordered selections, then divide out the orderings you have over-counted.
C(n,r) = P(n,r)/r! = n! / [r! (n-r)!]
2. Must-Know Rules
| Concept | Formula |
|---|---|
| Combinations of r from n | C(n,r) = n!/[r!(n-r)!] |
| Symmetry | C(n,r) = C(n, n-r) |
| Choose 0 or all | C(n,0) = C(n,n) = 1 |
| Choose 1 | C(n,1) = n |
| Choose 2 | C(n,2) = n(n-1)/2 |
| Total subsets of an n-set | 2^n |
The symmetry rule saves real time. C(10,8) = C(10,2) = 45. Always convert to the smaller r.
Values worth recognising:
| C(n,2) | n(n-1)/2 |
|---|---|
| C(5,2) | 10 |
| C(6,2) | 15 |
| C(8,2) | 28 |
| C(10,2) | 45 |
Handshake problems are always C(n,2): n people each shaking hands once with every other gives n(n-1)/2 handshakes.
Multiple groups. If you must choose from separate pools, multiply: choosing 2 from 5 men and 3 from 6 women gives C(5,2) x C(6,3) = 10 x 20 = 200.
"At least" problems. Either sum the individual cases, or subtract the complement from the total — the complement is usually faster.
3. Worked Examples
Example 1 — Basic committee
From 9 people, how many committees of 4 can be formed?
C(9,4) = 9!/(4! 5!) = (9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 3024/24 = 126.
Example 2 — Two pools
A team of 3 must include 2 engineers from 6 and 1 designer from 4. How many teams?
C(6,2) x C(4,1) = 15 x 4 = 60.
Example 3 — At least
From 5 men and 4 women, a committee of 3 is chosen. In how many ways can it include at least one woman?
Total committees = C(9,3) = 84.
All-men committees = C(5,3) = 10.
At least one woman = 84 - 10 = 74.
4. GRE Traps
- Using permutations for selections. A committee of 3 from 9 is 126, not P(9,3) = 504.
- Forgetting the symmetry shortcut and grinding C(12,10) the long way.
- Adding when you should multiply. Independent selections from separate pools multiply.
- Summing "at least" cases and missing one. Use the complement.
- Treating identical items as distinct (or vice versa). Read whether the objects are distinguishable.
- Double counting when a person could belong to two categories.
- Choosing "3 out of 9 in order" for a ranked list — that is a permutation. Re-read whether roles differ (president/secretary vs plain members).
5. Speed Tricks
- Order matters -> P. Order doesn't -> C. Decide before writing.
- Use the symmetry rule to make r as small as possible.
- C(n,2) = n(n-1)/2 covers handshakes, pairings, line segments between points, and diagonals.
- Cancel before multiplying in the factorial expression — never compute 9! outright.
- "At least one" -> total minus none.
- A quick sanity check: C(n,r) is always a whole number, and always less than or equal to P(n,r).
6. Self-Check
Q1. How many ways can 3 books be chosen from 7?
Q2. How many handshakes occur if 12 people each shake hands once with every other?
Q3. From 4 boys and 5 girls, how many groups of 4 contain exactly 2 boys?
Answers
A1. C(7,3) = (7x6x5)/6 = 35.
A2. C(12,2) = 12x11/2 = 66.
A3. C(4,2) x C(5,2) = 6 x 10 = 60.
7. One-Line Summary for the Board
Combination = selection without order. C(n,r) = C(n, n-r). Handshakes are always n(n-1)/2.