Week 12: Permutations & Combinations

1. The Fundamental Counting Principle

Before any formula, this one principle answers most simple counting questions directly: if a task has several independent steps, multiply the number of choices at each step.

the counting principle

If step 1 can be done in m ways, and step 2 (independently)
in n ways, the two together can be done in m x n ways.

Worked: a restaurant offers 4 starters, 6 mains and 3
desserts. How many different 3-course meals are possible?

  4 x 6 x 3 = 72 meals

Worked: how many 3-digit numbers can be formed using digits
0-9, if repetition is allowed and the first digit can't be 0?

  First digit: 9 choices (1-9, not 0)
  Second digit: 10 choices (0-9, repetition allowed)
  Third digit: 10 choices
  9 x 10 x 10 = 900

Most permutation and combination formulas are actually just this principle applied repeatedly and then simplified into a closed form — worth remembering when a formula doesn't quite fit an unusual problem, since falling back to the counting principle directly almost always still works.

2. Permutations vs. Combinations

One question decides which formula applies: does the order of selection matter? Arranging people in a row — order matters (permutation). Choosing a committee — order doesn't matter (combination).

the two formulas

Permutations (order MATTERS):
  nPr = n! / (n-r)!

Combinations (order DOESN'T matter):
  nCr = n! / (r! x (n-r)!)  =  nPr / r!

Worked: from 8 people, how many ways to choose a president,
secretary and treasurer (3 DIFFERENT roles -- order/role
matters)?
  8P3 = 8!/5! = 8x7x6 = 336

Worked: from 8 people, how many ways to choose a 3-person
committee (identical roles -- order doesn't matter)?
  8C3 = 8!/(3!x5!) = (8x7x6)/(3x2x1) = 336/6 = 56

Notice nCr = nPr / r! — a combination is exactly a permutation with the "which order were they picked in" information thrown away, divided out because every group of r people can be arranged in r! different orders, all of which count as the same single combination.

Ask "would swapping two selected items give a different result?"

If swapping two chosen items (two committee members, two selected fruits) produces a genuinely different outcome, it's a permutation. If it produces the exact same outcome (the same committee, the same set of fruits), it's a combination. This single test resolves the ambiguity faster than trying to remember which real-world scenarios "count as" each type.

3. Circular Arrangements

Arranging n distinct people around a circular table has fewer distinct arrangements than arranging them in a row — because rotating everyone by one seat produces an arrangement that's considered identical.

circular arrangement formula

Linear arrangement of n people: n!

Circular arrangement of n people: (n-1)!

Why: fix ONE person's position (to remove the "which seat is
seat 1" ambiguity, since a circle has no natural starting
point) -- the remaining (n-1) people can then be arranged in
(n-1)! ways relative to that fixed person.

Worked: in how many ways can 6 people be seated around a
circular table?
  (6-1)! = 5! = 120

A common variant: if the arrangement is also considered identical under reflection (flipping the whole circle, as with a necklace or bracelet, not a table where "left neighbor" and "right neighbor" are genuinely different), divide by an additional factor of 2: (n-1)!/2.

4. Grouping With Restrictions

"In how many ways can 5 boys and 3 girls be arranged in a row such that all 3 girls sit together?" Restrictions are handled by temporarily treating a restricted group as a single "super-item," then accounting for internal arrangement separately.

the "treat as one unit" technique

5 boys + 3 girls, girls must all sit TOGETHER.

Step 1: treat the 3 girls as ONE combined unit.
  Now arranging: 5 boys + 1 girl-unit = 6 "items"
  Arrangements of 6 items: 6! = 720

Step 2: within the girl-unit, the 3 girls can be arranged
among themselves in 3! = 6 ways

Total = 6! x 3! = 720 x 6 = 4,320

The reverse restriction ("no two girls sit together") uses a related but different technique — arrange the boys first (leaving gaps between and around them), then place the girls individually into those gaps, ensuring no two girls end up adjacent. Both techniques share the same principle: handle the restriction first, structurally, then count the remaining freedom separately.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Ask "does order matter?" before choosing a formula for every problem. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

Evaluate 5!.
Evaluate 4P2.
Evaluate 5C2.
In how many ways can 4 people be arranged in a row?
In how many ways can 3 books be chosen from 5 books?
In how many ways can a 3-digit code be formed from digits 1–9 without repetition?
Evaluate 6C6.
Evaluate 6C0.
How many ways can you arrange the letters of the word "CAT"?
If a coin is tossed 3 times, how many total outcomes are possible?

Easy — Answers

1) 120. 2) 4×3=12. 3) 10. 4) 4!=24. 5) 5C3=10. 6) 9×8×7=504. 7) 1. 8) 1. 9) 3!=6. 10) 2³=8.

Medium (11–20)

In how many ways can 5 different books be arranged on a shelf?
From 6 people, in how many ways can a committee of 3 be chosen?
From 8 people, in how many ways can a president and a vice-president be chosen?
In how many ways can the letters of the word "APPLE" be arranged?
How many 4-digit numbers can be formed using digits 1–6 without repetition?
In how many ways can 6 people be seated around a circular table?
From a group of 5 men and 3 women, in how many ways can a committee of 2 men and 1 woman be formed?
In how many ways can the letters of the word "BANANA" be arranged?
In how many ways can 3 prizes be distributed among 5 students, if each student can receive at most one prize?
In how many ways can 4 boys and 3 girls be arranged in a row such that all girls sit together?

Medium — Answers

11) 5!=120. 12) 6C3=20. 13) 8P2=56. 14) 5!/2!=60. 15) 6P4=360. 16) (6-1)!=120. 17) 5C2×3C1=30. 18) 6!/(3!×2!)=60 (3 A's, 2 N's). 19) 5P3=60. 20) Treat girls as one unit: 5!×3!=120×6=720.

Tough (21–30)

In how many ways can 7 people be seated around a circular table such that 2 particular people always sit together?
From a group of 6 men and 4 women, in how many ways can a committee of 5 be formed such that at least 3 men are included?
In how many ways can the letters of the word "MISSISSIPPI" be arranged?
How many 5-digit numbers greater than 40,000 can be formed using the digits 1, 2, 3, 4, 5 without repetition (using all five digits)?
In how many ways can a cricket team of 11 be selected from 15 players if 2 particular players must always be included?
In how many ways can 8 people be seated in a row such that 2 particular people never sit together?
Find the number of diagonals in a hexagon (6 sides) using combinations.
In how many ways can a committee of 4 be selected from 6 men and 4 women such that it includes at least 1 woman?
In how many ways can the digits 1, 2, 3, 4, 5 be arranged (using all 5) to form a number divisible by 2?
How many 4-letter words (not necessarily meaningful) can be formed from the letters of the word "NUMBER" without repetition?

Tough — Answers

21) Treat pair as 1 unit: (6-1)!×2!=120×2=240. 22) (6C3×4C2)+(6C4×4C1)+(6C5×4C0)=120+60+6=186. 23) 11 letters, M=1,I=4,S=4,P=2: 11!/(4!×4!×2!)=34,650. 24) First digit must be 4 or 5; remaining 4 digits arranged in 4!=24 ways each → 48 total. 25) Remaining 9 players choose 9 more (2 already included, 9 seats left from 13 remaining): 13C9=715. 26) Total 8!=40,320 minus together (2×7!=10,080) = 30,240. 27) nC2-n=6C2-6=15-6=9. 28) Total 10C4=210 minus all-men 6C4=15 → 195. 29) Last digit must be 2 or 4; remaining 4 digits arranged 4!=24 ways each → 48 total. 30) NUMBER has 6 distinct letters: 6P4=360.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Why is choosing a president, secretary and treasurer from 8 people a permutation problem, while choosing a 3-person committee is a combination problem?

Swapping two people between the president and secretary roles produces a genuinely different outcome (different people in different jobs), so which position each person fills matters — that's a permutation. Swapping two people's order within a committee doesn't change who's on the committee at all, since the roles are identical, which makes it a combination.

Why does nCr equal nPr divided by r! specifically, rather than some other number?

Every distinct group of r items can be internally arranged in exactly r! different orders, and nPr counts each of those orderings as a separate outcome. Since a combination treats all r! of those orderings as the same single group, dividing nPr by r! collapses each group of r! duplicate permutations down to the one combination they all represent.

Why does a circular arrangement of n people use (n-1)! instead of n!?

A circle has no fixed starting point, so rotating every person by one seat produces an arrangement that's considered identical to the original — n! would count each of these n rotations of the same arrangement as separate outcomes. Fixing one person's position eliminates that rotational duplication, leaving only (n-1)! genuinely distinct arrangements for the remaining people.

Why does the "treat as one unit" technique multiply by the internal arrangements of the grouped items, rather than just counting the group as a single fixed block?

Treating the group as one unit only accounts for where the group sits relative to everything else — it says nothing about the order of the distinct people within that group, which is a separate, independent choice. Multiplying by the internal arrangements (r! for r grouped people) accounts for both facts simultaneously: where the group is, and how its members are ordered inside it.