4.9 Permutations & Combinations
Key Takeaways
- Permutation & Combination is a named bullet of the CIL Paper-I Quantitative Aptitude syllabus, and the whole topic turns on whether order matters.
- A permutation counts arrangements where order matters and equals n factorial divided by (n minus r) factorial, while a combination counts selections and divides that further by r factorial.
- Arranging n distinct objects in a circle gives (n minus 1) factorial arrangements because rotations of the same arrangement are not distinct.
- When letters repeat, the number of distinct arrangements is n factorial divided by the product of the factorials of each repetition count.
The Two Counting Principles
Multiplication principle. If a first task can be done in $m$ ways and, for each of those, a second in $n$ ways, the pair can be done in $m \times n$ ways. Use it when tasks happen in sequence — signalled by the word and.
Addition principle. If a task can be done in $m$ ways or in $n$ ways, with no overlap, the total is $m + n$. Use it for mutually exclusive alternatives — signalled by the word or.
Factorials
The convention $0! = 1$ is not arbitrary; it is what makes the formulae below work at their boundaries, and it appears as an objective item in its own right.
| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| $n!$ | 1 | 1 | 2 | 6 | 24 | 120 | 720 | 5040 |
Permutations: Order Matters
The number of ways of arranging $r$ objects chosen from $n$ distinct objects is
Example. In how many ways can a first, second and third prize be awarded among 6 candidates? Order matters because the prizes differ:
Combinations: Order Does Not Matter
The number of ways of selecting $r$ objects from $n$ is
Example. In how many ways can a committee of 3 be chosen from 6 people?
Note that $120 = 20 \times 3!$ — each selection of three can be arranged in $3! = 6$ orders, which is precisely the difference between the two counts.
Properties of combinations
The first identity is a genuine time-saver: computing $^{20}C_{18}$ as $^{20}C_{2} = 190$ takes seconds.
The Deciding Question
| Signal words | Use |
|---|---|
| arrange, order, rank, seat, form a number, assign distinct posts | Permutation |
| select, choose, committee, team, group, handshake, draw | Combination |
A committee is a combination; a committee with designated posts such as chairman and secretary is a permutation, or a combination followed by an arrangement.
Arrangements with Repetition
When some objects are identical, divide by the factorial of each repetition count:
Example. How many distinct arrangements of the letters of ENGINEER?
The word has 8 letters with E appearing 3 times and N appearing 2 times:
Circular Arrangements
Seating $n$ distinct people around a round table gives
arrangements, because rotating everyone by one seat does not create a new arrangement — only relative order matters. Fix one person's position and arrange the remaining $n-1$.
If clockwise and anticlockwise arrangements are not distinguished, as with a garland of flowers or a necklace of beads, halve the result:
Example. Six engineers sit around a circular table: $(6-1)! = 120$ ways. If instead six identical-looking beads form a necklace: $5!/2 = 60$.
Restricted Arrangements
Objects that must stay together
Treat the block as a single unit, arrange the units, then arrange within the block.
Example. In how many ways can 5 people be arranged in a row if two particular people must sit together? Treat the pair as one unit, giving 4 units arranged in $4! = 24$ ways, and the pair internally in $2! = 2$ ways. Total $= 48$.
Objects that must be separated
Count the total and subtract the together case.
Example. With the same 5 people, the number of arrangements in which the two are not together is $5! - 48 = 120 - 48 = 72$.
Fixed positions
Example. How many 4-digit even numbers can be formed from the digits 1 to 7 without repetition? The units digit must be 2, 4 or 6, giving 3 choices, and the remaining three positions are filled from the other 6 digits in $^{6}P_{3} = 120$ ways. Total $= 3 \times 120 = 360$.
Selection Problems
Example. From 7 men and 5 women, how many committees of 5 can be formed with exactly 3 men?
Example. How many committees of 5 contain at least 3 men? Sum the cases of exactly 3, 4 and 5 men:
The phrase at least always means summing cases, or subtracting the complement from the total when that is shorter.
Standard Results Worth Recall
| Situation | Count |
|---|---|
| Handshakes among $n$ people | $^{n}C_{2}$ |
| Diagonals of an $n$-sided polygon | $^{n}C_{2} - n$ |
| Triangles from $n$ points, no three collinear | $^{n}C_{3}$ |
| Subsets of a set of $n$ elements | $2^{n}$ |
| Ways to distribute $n$ distinct objects into $r$ distinct boxes | $r^{n}$ |
All of these appear regularly, and each takes a few seconds once recognised.
The number of distinct arrangements of the letters of the word ENGINEER is:
The number of ways in which 6 distinct people can be seated around a circular table is:
From 7 men and 5 women, the number of committees of 5 members containing exactly 3 men is:
Which situation calls for a combination rather than a permutation?