Order is the whole game. If rearranging the same items counts as a new outcome (seats, ranks, codes), it is a permutation. If only the group matters (a committee, a hand of cards), it is a combination.
The two engines
nPr = n! / (n−r)! for arrangements where order matters; nCr = n! / (r!(n−r)!) for selections where only the group matters. Symmetry: nCr = nC(n−r).
Probability
P(event) = favourable outcomes / total outcomes — always a value between 0 and 1.
How to Approach It
- Ask whether order matters — rearranging counts as new (seats, ranks, codes) → permutation; only the group matters (team, hand of cards) → combination. This single decision picks your formula.
- Handle repetition deliberately — when items repeat, divide by the factorial of each repeat count; when repeated selections are allowed, the counting rule changes, so check.
- For probability, count both totals — favourable ÷ total, counted carefully. Your answer must land between 0 and 1 — if not, the total is wrong.
- Switch to the complement for 'at least' — compute 1 minus P(none); it is almost always faster than summing cases.
Techniques & Methods
- Order test — order matters → permutation; only the group matters → combination. e.g. ranking → P; choosing a committee → C.
- Divide by repeats — arrangements with repeated items = n! ÷ (factorial of each repeat). e.g. LEVEL → 5!/(2!2!) = 30.
- Use the complement — P(at least one) = 1 − P(none). e.g. at least one head in two tosses = 1 − 1/4 = 3/4.
- nCr symmetry — nCr = nC(n−r); pick the smaller r to cut the work. e.g. 20C18 = 20C2 = 190.
The Edge
When letters repeat, divide by the factorial of each repeat count. LEVEL has 5 letters with L,E twice each: arrangements = 5!/(2!×2!) = 30, not 120. Use nCr = nC(n−r) to shrink the arithmetic — 20C18 is just 20C2 = 190.Worked example
In how many ways can the letters of MANGO be arranged?
- First decide the type: arranging letters in order is a permutation, and here all five letters are different with no repeats.
- The number of arrangements of n distinct items is n factorial (n!).
- With n = 5: 5! = 5 × 4 × 3 × 2 × 1.
- Multiply through: that equals 120 distinct arrangements.
Answer: 120 arrangements
Worked example
A bag has 4 red and 3 blue balls. One ball is drawn. What is the probability it is red?
- Probability is favourable outcomes ÷ total outcomes, so count both sides first.
- Total balls = 4 red + 3 blue = 7, which is the size of the sample space.
- Favourable outcomes (drawing a red) = 4.
- So P(red) = 4/7, which sensibly lies between 0 and 1.
Answer: P(red) = 4/7
Worked Drills
Worked example
How many 4-digit numbers can be formed from the digits 1-7 without repetition?
- Order matters and no repeats, so this is 7P4 = 7 × 6 × 5 × 4.
- That equals 840.
Answer: b) 840
Worked example
A committee of 3 men and 2 women is formed from 6 men and 5 women. The number of ways is:
- Only the group matters → combinations. Choose 3 men from 6 and 2 women from 5.
- C(6,3) × C(5,2) = 20 × 10 = 200.
Answer: b) 200
Worked example
The number of distinct arrangements of the letters of MISSISSIPPI is:
- 11 letters with repeats: I four times, S four times, P twice.
- Arrangements = 11!/(4!4!2!) = 34650.
Answer: a) 34650
Worked example
Two dice are thrown. The probability that the sum is a multiple of 4 is:
- Multiples of 4 are sums 4, 8 and 12, with 3 + 5 + 1 = 9 favourable outcomes.
- Total outcomes = 36, so P = 9/36 = 1/4.
Answer: a) 1/4
Worked example
Two cards are drawn from a pack without replacement. The probability both are aces is:
- First ace: 4/52. After removing it, second ace: 3/51.
- Multiply: (4/52)(3/51) = 1/221.
Answer: a) 1/221
Worked example
The number of distinct arrangements of the letters of LEADER is:
- 6 letters with E repeated twice; all others distinct.
- Arrangements = 6!/2! = 360.
Answer: b) 360
Worked example
The number of arrangements of the letters of LEVEL is:
- 5 letters with L twice and E twice.
- Arrangements = 5!/(2!2!) = 120/4 = 30.
Answer: b) 30
Worked example
The probability of a sum of 7 when two dice are rolled is:
- Favourable pairs giving 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 outcomes.
- P = 6/36 = 1/6.
Answer: a) 1/6
⚠ Watch out
- "Arrange" / "order" / "rank" → permutation; "select" / "choose" / "committee" → combination.
- Repeated letters need division by their factorials.
- Probability can never exceed 1 — if your answer does, the total is wrong.
Takeaways
- Settle "does order matter?" first — it picks permutation vs combination instantly.
- Repeated items mean dividing by the factorial of each repeat count.
- Use nCr = nC(n−r) to choose the smaller r and shrink the arithmetic.
- For "at least one", compute 1 − P(none) instead of summing cases.