Counting, Probability and Set Theory

Two or three of the twenty-two QA questions come from this block, and they are the questions candidates most often get wrong while feeling certain. The arithmetic is trivial — the difficulty is entirely in deciding what to count. Does order matter here? Are repetitions allowed? Is this "at least one" or "exactly one"? Get the model right and the calculation takes ten seconds; get it wrong and you will produce a confident wrong answer. So this chapter spends its effort on the decisions rather than on the formulas: four questions to ask before counting anything, the complement trick that turns most hard probability questions into easy ones, and the two-set and three-set formulas that make Venn questions mechanical.

Four Questions to Ask Before Counting

Almost every counting mistake is a modelling mistake made in the first ten seconds. Asking these four questions in order removes most of them, and costs less time than the recovery does.

Does order matter? If yes it is a permutation (arrangements, rankings, seating, forming a number, choosing a president and a secretary). If no it is a combination (committees, teams, handshakes, selecting fruit). "Selecting 3 from 8" is 56; "arranging 3 from 8" is 336, six times more, because each selection can be ordered in 3! = 6 ways.
Are repetitions allowed? A four-digit PIN allows them (10⁴ = 10,000 possibilities); a four-digit number with distinct digits does not. Watch the first digit: a four-digit number cannot start with 0, so the count is 9 × 9 × 8 × 7 = 4536 and not 10 × 9 × 8 × 7.
Are the objects distinguishable? Five different books arranged on a shelf give 5! = 120 orders. Five identical books give one. The word MISSISSIPPI has 11 letters but only 11!/(4!·4!·2!) = 34,650 distinct arrangements, because the repeated letters cannot be told apart.
Is it easier to count the opposite? "At least one" almost always is. The number of ways to get at least one head in five tosses is 2⁵ − 1 = 31, counted as "all outcomes minus the one with no heads". Counting the cases directly means adding five terms.

⚠️ "At least" and "exactly" are different questions

Read those two words as instructions, because paper-setters put both answers among the options. "At least two" includes two, three, four and more; "exactly two" excludes all of them. If the phrasing is "at least", reach for the complement. If it is "exactly", count that case and only that case.

The Counting Tools

What to use, and when
Tool	Formula and use
Combination	C(n, r) = n!/[r!(n − r)!] — selections where order is irrelevant
Permutation	P(n, r) = n!/(n − r)! — arrangements, which are selections then ordered
Repeated letters	n! divided by the factorial of each repeat count
Circular arrangements	(n − 1)! — one seat is fixed because only relative position matters
Items that must stay together	Glue them into one block, arrange, then arrange within the block
Items that must stay apart	Arrange the others first, then place these in the gaps between them
Identical items into distinct groups	C(n + r − 1, r − 1) — the "stars and bars" count

Worked example (together). In how many ways can 5 people be seated in a row if two particular people must sit together? Treat the pair as one block: 4 objects arrange in 4! = 24 ways, and the pair can swap internally in 2 ways, giving 24 × 2 = 48. Out of 5! = 120 total, so they sit together in 40% of arrangements — which is a useful sanity check, since intuition says "less than half".
Worked example (apart). In how many ways can 5 people be seated so that two particular people are never together? Total minus together: 120 − 48 = 72. The complement again — counting the "never" cases directly would need cases for every gap.
Worked example (committee with a condition). From 6 men and 4 women, how many committees of 4 contain at least one woman? Total committees C(10,4) = 210; all-male committees C(6,4) = 15; so 210 − 15 = 195. Adding the four cases (1, 2, 3 or 4 women) gives 80 + 90 + 24 + 1 = 195 as well — same answer, four times the work.
Worked example (stars and bars). In how many ways can 10 identical chocolates be given to 3 children if a child may get none? C(10 + 3 − 1, 3 − 1) = C(12, 2) = 66. If every child must get at least one, hand out one each first and distribute the remaining 7: C(9, 2) = 36.

Probability — Count Twice, Divide Once

Probability at this level is counting with a division at the end: favourable outcomes ÷ total outcomes, provided every outcome is equally likely. That proviso is where questions hide their difficulty, so state your sample space explicitly before dividing.

The complement rule. P(at least one) = 1 − P(none). Two dice thrown: the chance of at least one six is 1 − (5/6)² = 1 − 25/36 = 11/36. Note it is not 1/6 + 1/6 = 1/3 — that double-counts the outcome where both are sixes, and 11/36 < 12/36 by exactly that one outcome.
Addition, with the overlap removed. P(A or B) = P(A) + P(B) − P(A and B). Drawing one card: P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. The king of hearts is one card and must be counted once.
With and without replacement. Two cards drawn: P(both kings) is (4/52)(4/52) = 1/169 with replacement, but (4/52)(3/51) = 1/221 without. CAT nearly always means without replacement, and the phrase to look for is "one after another without replacement" — or simply "two cards are drawn", which means the same thing.
Conditional probability. P(A given B) = P(A and B)/P(B). A die shows an even number; what is the chance it is a 6? The condition shrinks the sample space to {2, 4, 6}, so the answer is 1/3, not 1/6. Re-reading the question as "the sample space is now smaller" is faster than the formula.
A probability is never outside 0 and 1. It sounds too obvious to check, and it catches real errors: if a chain of multiplications returns 7/5, a "without replacement" was treated as "with", or two probabilities were added that should have been multiplied.

Set Theory — Two Formulas and a Diagram

Two sets. n(A ∪ B) = n(A) + n(B) − n(A ∩ B). In a class of 50 where 30 play cricket, 25 play football and 10 play both, the number playing at least one is 30 + 25 − 10 = 45, so 5 play neither. That last step — the "neither" count — is what the question usually asks for.
Three sets. n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A∩B) − n(B∩C) − n(C∩A) + n(A∩B∩C). Note the signs: pairs subtract, the triple adds back. Getting that final plus wrong is the standard error, and it is worth remembering why — the element in all three was removed three times by the pairwise subtractions, so one must be returned.
Maximum and minimum of an overlap. These are pure reasoning, and CAT prefers them to plain substitution. If 70% of a group like tea and 80% like coffee, the overlap is at most 70% (everyone who likes tea could also like coffee) and at least 50% (70 + 80 − 100). The lower bound is the one people miss: two majorities in a group of 100 must overlap by at least their sum minus 100.
Draw the diagram when there are three sets. Fill the centre first, then the pairwise regions (subtracting the centre from each), then the single regions. Working outward-in is how people end up counting the middle three times.

🎯 Where this block reappears

Set counting is not only a QA topic. A DILR set built on "how many customers bought at least two of the three products" is the same three-set formula with a table instead of a Venn diagram. Time spent here pays twice.

Key takeaways

Order matters → permutation; order does not → combination. Decide this before touching a formula, because everything after it depends on the answer.
For "at least one", count the complement: total minus the none case. Direct counting needs several cases and each is a chance to slip.
A four-digit number cannot start with zero. Read every counting question for the constraint hidden in an ordinary word like "number" or "distinct".
Items that must stay together become one block; items that must stay apart go into the gaps left by the others.
Circular arrangements of n objects are (n − 1)!, because only relative position matters — fix one seat and arrange the rest.
In the three-set formula the pairwise overlaps subtract and the triple overlap adds back. The element in all three was removed three times, so one must be returned.
Two majorities in a group of 100 must overlap by at least their sum minus 100. That lower bound is the half of a max/min question candidates forget.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

How many four-digit numbers can be formed using the digits 0 to 9 without repeating a digit?
1. 5040
2. 4536
3. 10000
4. 4500
Show answer
Answer: B — 4536
The first digit cannot be 0, so it has 9 choices. The remaining three positions can use any unused digits, including 0: 9, then 8, then 7. The count is 9 × 9 × 8 × 7 = 4536. Option A, 5040, is 10 × 9 × 8 × 7 — the answer you get by allowing a leading zero, which would make it a three-digit number.
In how many ways can 5 people be seated in a row if two particular people must sit next to each other?
1. 24
2. 48
3. 72
4. 120
Show answer
Answer: B — 48
Treat the pair as a single block. Four objects (the block plus three people) arrange in 4! = 24 ways, and within the block the two can swap in 2 ways: 24 × 2 = 48. Option A forgets the internal swap; option C, 72, is the complement — the arrangements where they are not together (120 − 48).
From 6 men and 4 women, a committee of 4 is to be formed with at least one woman. The number of such committees is:
1. 195
2. 210
3. 80
4. 15
Show answer
Answer: A — 195
Use the complement. All committees of 4 from 10 people: C(10,4) = 210. Committees with no woman are chosen entirely from the 6 men: C(6,4) = 15. So 210 − 15 = 195. Checking the long way: exactly 1 woman C(4,1)C(6,3) = 4 × 20 = 80; exactly 2, C(4,2)C(6,2) = 6 × 15 = 90; exactly 3, C(4,3)C(6,1) = 4 × 6 = 24; exactly 4, C(4,4) = 1. Total 80 + 90 + 24 + 1 = 195 ✓ Option C is the "exactly one woman" count, which the wording does not ask for.
The number of distinct arrangements of the letters of the word MISSISSIPPI is:
1. 39916800
2. 34650
3. 11550
4. 69300
Show answer
Answer: B — 34650
The word has 11 letters: M once, I four times, S four times, P twice. Identical letters cannot be told apart, so divide 11! by 4! for the Is, 4! for the Ss and 2! for the Ps: 39,916,800 / (24 × 24 × 2) = 39,916,800 / 1152 = 34,650. Option A is 11! itself, the count if every letter were distinguishable.
Two dice are thrown. The probability of getting at least one six is:
1. 1/3
2. 11/36
3. 1/36
4. 25/36
Show answer
Answer: B — 11/36
Use the complement: P(no six) = (5/6)(5/6) = 25/36, so P(at least one six) = 1 − 25/36 = 11/36. Option A, 1/3 = 12/36, is what you get by adding 1/6 + 1/6 — which counts the double-six outcome twice, and the difference between 12/36 and 11/36 is exactly that one outcome.
Two cards are drawn one after another from a standard pack of 52, without replacement. The probability that both are kings is:
1. 1/169
2. 1/221
3. 1/13
4. 4/52
Show answer
Answer: B — 1/221
The first card is a king with probability 4/52. Given that, only 3 kings remain among 51 cards, so the second is a king with probability 3/51. Multiplying: (4/52)(3/51) = 12/2652 = 1/221. Option A, 1/169, is (4/52)² — the answer with replacement, which the question explicitly excludes.
In a class of 50 students, 30 play cricket, 25 play football and 10 play both. How many play neither game?
1. 5
2. 10
3. 45
4. 0
Show answer
Answer: A — 5
The number playing at least one game is 30 + 25 − 10 = 45, subtracting the 10 who were counted in both totals. So 50 − 45 = 5 play neither. Option C, 45, is the "at least one" count — the intermediate step, and a frequent distractor because the question asks for what is left over.
In a survey, 70% of people said they like tea and 80% said they like coffee. The minimum possible percentage who like both is:
1. 70%
2. 50%
3. 10%
4. 0%
Show answer
Answer: B — 50%
At most 100% of people can like at least one drink, so n(tea) + n(coffee) − n(both) ≤ 100, that is 70 + 80 − n(both) ≤ 100, giving n(both) ≥ 50. The overlap is therefore at least 50%, and it reaches exactly 50% when everybody likes at least one. Option D would require 150% of people to exist. (The maximum, for contrast, is 70% — everyone who likes tea could also like coffee.)