OMPT drill — Probability & statistics

OMPT counting and combinatorics practice: lesson and 10 problems

Permutations versus combinations: does order matter? Answer that one question correctly and the formulas pick themselves. Write the formula down before plugging in numbers; on this strand the setup is the whole battle.

Tested inOMPT-E

OMPT counting and combinatorics: the lesson

What the OMPT tests in counting and combinatorics

Combinatorics is its own chapter on OMPT-E and appears nowhere else. The questions are short word problems: how many codes, how many committees, how many routes, how many seatings with a condition attached. Behind every one of them sit three ideas. The multiplication principle (stages multiply), permutations (arrangements, where order counts) and combinations (selections, where it does not). A fourth idea, arrangements of objects with repeats, turns up about once per paper, and a fifth, counting the complement for "at least one" questions, is the usual hardest item.

The chapter also feeds the probability questions that follow it: a probability of the form favourable outcomes over total outcomes is two counting problems in disguise. So a weak count costs marks twice. In my sessions the students who struggle here are rarely bad at arithmetic. They pick the wrong formula because they never asked the two questions that decide everything: does order matter, and can things repeat?

Notation and answer format the OMPT expects

Answers are whole numbers typed into the box: 56, not $(3 8)$ . You must finish the arithmetic, so compute binomial coefficients by cancelling before multiplying.
The exam writes combinations as $(k n)$ ; some textbooks write $n C r$ or $C (n, k)$ . All three mean $\frac{n !}{k ! ( n - k )!}$ .
Permutations of $k$ from $n$ are the truncated product $n (n - 1) \dots (n - k + 1)$ , which equals $\frac{n !}{( n - k )!}$ .
Factorials grow fast. $10! = 3628800$ . You will not be asked to multiply that out; a question that seems to need it is asking you to cancel.

Worked example 1: podium versus squad

Eight runners. In how many ways can gold, silver and bronze be awarded? And in how many ways can a relay squad of three be picked?
The podium has three labelled slots. Gold has 8 candidates, silver 7, bronze 6, so $8 \cdot 7 \cdot 6 = 336$ . Order matters because swapping gold and silver changes the result.
A squad has no labels. Each squad of three people appears $3! = 6$ times among the 336 podiums, once per internal ordering, so the number of squads is $\frac{336}{6} = 56$ .
That division is exactly what $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}$ computes. Same numbers, one extra step, and the step is decided by the word "squad".

Worked example 2: codes with and without repetition

How many 4-digit codes can be made from the digits 0 to 9, first if digits may repeat, then if they may not?
With repetition every slot keeps all 10 options: $1 0^{4} = 10000$ .
Without repetition the options shrink: $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ .
The repetition question is the second thing to settle, and OMPT-E alternates between the two on purpose. Note the without-repetition count is roughly half the other, which is the arithmetic behind the birthday problem.

The full lesson covers arrangements with repeated letters, committees with conditions, seatings where two people must or must not sit together, grid routes through a fixed corner, subsets, and the complement trick for "at least one". It ends with the edge cases the exam likes and a list of the exact slips I mark most often.

How each OMPT variant tests counting and combinatorics

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Not tested
OMPT-A has no combinatorics chapter. Spend the time on exponential functions and logarithms instead.
OMPT-B
64 questions
Not tested
OMPT-B stops at integration and never asks a counting question; the probability strand is OMPT-E only.
OMPT-C
30 questions
Not tested
OMPT-C covers numbers, algebra, linear equations and geometry. No counting.
OMPT-D
21 questions
Not tested
OMPT-D is the calculus and trigonometry paper. Counting is not in its 11 chapters.
OMPT-E
40 questions
Testedroughly 3 to 4 of 40 questions, plus the counting hidden inside probability questions
Official chapter: Combinatorics, 10% of the paper.
Short word problems with a whole-number answer: codes, committees, seatings, routes. One item per paper usually needs the complement ("at least one") or a repeated-letter correction.
OMPT-F
21 questions
Not tested
OMPT-F is business calculus with six chapters and no statistics. Skip this topic.
OMPT-G
26 questions
Not tested
OMPT-G goes into sequences, trig, geometry, mechanics and chemistry, not counting.

OMPT counting and combinatorics practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

A study group of 3 students must be chosen from a class of 8. How many different groups are possible?

Show the worked solution

Answer: 56

Order does not matter in a group, so I use a combination: $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = \frac{336}{6} = 56$ .

Problem 2Multiple choice

How many distinct arrangements are there of the letters of KOFFIE?

A $720$
B $360$
C $120$
D $240$

Show the worked solution

Answer: B — $360$

KOFFIE has 6 letters with the F appearing twice. Total orderings of 6 letters is $6! = 720$ , but swapping the two identical F's gives the same word, so I divide by $2!$ : $\frac{720}{2} = 360$ .

Problem 3Numeric answer

A door code consists of 3 different digits chosen from 1 through 7, entered in order. How many codes are possible?

Show the worked solution

Answer: 210

Order matters and repetition is banned, so I multiply the shrinking choices: $7 \cdot 6 \cdot 5 = 210$ .

Problem 4Multiple choice

Five friends sit in a row for a photo. Two of them insist on sitting next to each other. How many seatings are possible?

A $120$
B $24$
C $48$
D $96$

Show the worked solution

Answer: C — $48$

I glue the inseparable pair into one block, leaving 4 objects to arrange: $4! = 24$ ways. Inside the block the two friends can swap: another factor 2. Total: $24 \cdot 2 = 48$ .

Problem 5Spot the error

A student counts how many different pairs of delegates can be picked from 10 councillors. Step 1: there are 10 choices for the first delegate. Step 2: there are 9 choices for the second, giving $10 \cdot 9 = 90$ . Step 3: so there are 90 possible pairs. Which step contains the error?

Show the worked solution

Answer: Step 3

Steps 1 and 2 correctly count ordered selections: 90 ordered pairs exist. The mistake is Step 3's conclusion that these are the pairs. Picking Anna then Bram and picking Bram then Anna form the same pair, so every pair was counted twice. The number of pairs is $\frac{90}{2} = (2 10) = 45$ .

Problem 6Numeric answer

On a city grid, you walk from corner $A$ to corner $B$ by going exactly 4 blocks east and 3 blocks north, in any order. How many different routes are there?

Show the worked solution

Answer: 35

Every route is a sequence of 7 steps of which 3 are "north". Choosing which positions the north steps occupy gives $(3 7) = \frac{7 \cdot 6 \cdot 5}{6} = 35$ routes. Choosing the east positions instead, $(4 7)$ , gives the same count.

Problem 7Multiple choice

How many distinct arrangements are there of the letters of PARALLEL?

A $3360$
B $40320$
C $6720$
D $1680$

Show the worked solution

Answer: A — $3360$

PARALLEL has 8 letters: one P, two A's, one R, three L's and one E. If all eight were different there would be $8! = 40320$ orderings. Swapping the identical A's or the identical L's changes nothing, so I divide by $2!$ for the A's and $3!$ for the L's: $\frac{40320}{2 \cdot 6} = \frac{40320}{12} = 3360$ . Quick check on the divisor: $2! \cdot 3! = 12$ , and $3360 \cdot 12 = 40320$ .

Problem 8Numeric answer

A committee of 5 is chosen from 7 women and 5 men. How many committees contain exactly 3 women?

Show the worked solution

Answer: 350

Exactly 3 women means 3 women and 2 men, chosen independently, so I multiply the two selections: $(3 7) \cdot (2 5)$ . The first is $\frac{7 \cdot 6 \cdot 5}{6} = 35$ , the second is $\frac{5 \cdot 4}{2} = 10$ , giving $35 \cdot 10 = 350$ . As a plausibility check, all committees of 5 from 12 number $(5 12) = 792$ , and 350 is a sensible slice of that.

Problem 9Numeric answer

A password has 5 characters, each chosen from the letters A, B, C, D, E, and letters may repeat. How many such passwords contain at least one A?

Show the worked solution

Answer: 2101

I count the complement. Every password: 5 options per slot, $5^{5} = 3125$ . Passwords with no A at all use only 4 letters per slot: $4^{5} = 1024$ . So at least one A: $3125 - 1024 = 2101$ . Counting directly (exactly one A, exactly two, and so on) gives the same total but takes five cases.

Problem 10Numeric answer

A club can form a committee of 2 from its members in exactly 66 ways. In how many ways can the same club form a committee of 3?

Show the worked solution

Answer: 220

First I recover the number of members $n$ from $(2 n) = 66$ , which is $\frac{n ( n - 1 )}{2} = 66$ , so $n^{2} - n - 132 = 0$ . Factoring: $(n - 12) (n + 11) = 0$ , and only $n = 12$ makes sense. Then $(3 12) = \frac{12 \cdot 11 \cdot 10}{6} = \frac{1320}{6} = 220$ . Check the first step: $(2 12) = 66$ .

OMPT counting and combinatorics: questions students ask

Is combinatorics on the OMPT-A or OMPT-B?

No. Combinatorics is a chapter of OMPT-E only. OMPT-A, B, C, D, F and G do not contain a counting chapter, and their syllabi list nothing about permutations or combinations. If your programme asks for OMPT-A or OMPT-B, you can leave this topic out entirely and put the time into functions and differentiation.

How many combinatorics questions are on the OMPT-E?

The official syllabus lists Combinatorics as one of twelve chapters on a 40-question paper, so my estimate is three to four direct counting questions. Counting also hides inside the probability questions, where a probability is favourable outcomes divided by total outcomes, so the skill is worth more than its own chapter suggests.

Permutation or combination: how do I tell which one the OMPT wants?

Ask whether swapping two of the chosen items produces a different outcome. Gold and silver medals, a code typed in order, seats in a row: yes, so multiply the shrinking choices (a permutation). A committee, a squad, a hand of cards: no, so divide those ordered counts by the number of orderings of the chosen items (a combination).

Do I need a calculator for counting questions on the OMPT?

No, and the questions are built so you do not. Binomial coefficients are meant to be cancelled before multiplying: 9 choose 4 is (9 times 8 times 7 times 6) over 24, which cancels to 126 in two lines. If a count seems to need a huge factorial written out, you have skipped the cancelling step.

How do I count "at least one" on the OMPT?

Count the complement. Work out every outcome, then subtract the outcomes with none of the wanted item. A 4-digit code with at least one 7 is 10 to the power 4 minus 9 to the power 4, which is 3439. Counting the cases with exactly one, two, three and four sevens gives the same number but takes four times the work and invites double counting.

Test hubs that cover counting and combinatorics:OMPT-E All 28 topics

OMPT counting and combinatorics practice: lesson and 10 problems

OMPT counting and combinatorics: the lesson

What the OMPT tests in counting and combinatorics

Notation and answer format the OMPT expects

Worked example 1: podium versus squad

Worked example 2: codes with and without repetition

How each OMPT variant tests counting and combinatorics

OMPT counting and combinatorics practice problems

OMPT counting and combinatorics: questions students ask

Related OMPT topics and test hubs