IOQM-level practice

IOQM combinatorics practice set

8 original problems at the level of the first olympiad stage: counting with choices, gaps, stars and bars, inclusion–exclusion and pigeonhole. Each answer is a whole number. Try each one before opening the hints; the full solution explains why the method works and where the idea leads.

The first 3 are free with full solutions. Hints, answer checking and full solutions for problems marked ‘With a plan’ are included with every A Level, IB, IGCSE and CBSE plan. See plans. These are our own problems, written for this site; none is a past IOQM or RMO question or a reworded one.

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Problem S160

CombinatoricsShort answer

How many four-digit numbers have digits that strictly increase from left to right (for example 1359)?

Hint

Choose the digits first. In how many orders can they be written?

Second hint

0 cannot be used (it would have to come first). Choose 4 digits from 1 to 9.

Full worked solution

Answer: 126

0 can never appear: it would be the smallest digit and so the leading digit.
Any 4 different digits from 1 to 9 can be written in increasing order in exactly one way.
So the count is the number of 4-element subsets of {1, …, 9}: C(9, 4).
C(9, 4) = 126.

Why this works: When the order is forced, an arrangement is just a choice of a set.

Where it leads: This bijection between increasing sequences and subsets is the basis of many counting arguments, including stars and bars.

Strategy: Organised cases

Problem S161

CombinatoricsShort answer

In how many ways can 10 identical sweets be shared among 4 children so that every child gets at least one?

Hint

Give each child one sweet first.

Second hint

Then share the remaining 6 freely: stars and bars with 6 stars and 3 bars.

Full worked solution

Answer: 84

Give one sweet to each child; 6 sweets remain, to be shared with no restriction.
A sharing is a row of 6 sweets and 3 dividers (stars and bars).
Choose the positions of the 3 dividers among 9 places: C(9, 3).
C(9, 3) = 84.

Why this works: Pre-allocating the minimum turns an ‘at least one’ count into an unrestricted one.

Where it leads: Stars and bars counts solutions of x₁ + … + x_k = n and underlies generating functions.

Strategy: Working backwards

Problem S162

CombinatoricsShort answer

The letters of the word PEPPER are arranged in a row. In how many arrangements are no two P’s next to each other?

Hint

Arrange the other letters first, then place the P’s in the gaps.

Second hint

E, E, R can be arranged in 3 ways; they create 4 gaps for the three P’s.

Full worked solution

Answer: 12

Arrange E, E, R: 3!/2! = 3 ways.
Each arrangement leaves 4 gaps (ends included): _ E _ E _ R _.
Place the three identical P’s in 3 different gaps: C(4, 3) = 4 ways.
Total: 3 × 4 = 12.

Why this works: Placing the restricted letters into gaps between the others guarantees they are never adjacent.

Where it leads: The gap method also counts seatings where certain people must not sit together, and binary strings with no two adjacent 1s.

Strategy: Organised cases