OMPT drill — Probability & statistics

Counting and combinatorics

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

Lesson

Every counting problem on the OMPT reduces to one question asked early: does order matter? Arranging five people in a row cares about order, swapping two of them creates a new arrangement. Choosing a committee of three does not: the committee is the same regardless of who was named first. Answer the order question and the formula picks itself: arrangements use permutations, selections use combinations.

Underneath both formulas sits something simpler, the multiplication principle: when a task happens in stages, the total count is the product of the options at each stage. Three shirts and four trousers make $3 \times 4 = 12$ outfits. Most exam questions can be solved directly from this principle, filling slots one at a time, and I would rather see a student reason through slots correctly than quote a memorised formula at the wrong situation. The formulas are compressed slot-reasoning; keep the reasoning available for when a question does not fit the standard moulds.

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.