Counting and Probability at CMAT Register

CMAT's Quantitative Techniques & Data Interpretation section is 20 questions at +4 and −1, and one or two of them come from counting and probability. That ratio is the first thing to know, because it decides how much of your preparation this deserves: enough to answer a clean question in under a minute, not enough to justify a month on restricted arrangements. The good news is that CMAT keeps this block clean — factorials, one permutation or combination, one straightforward probability, occasionally a repeated-letter arrangement. There is no advanced conditional probability, no Bayes, no inclusion-exclusion beyond two events. Almost every question in this chapter is decided by one prior question — does the order matter? — and by one habit: computing the complement when the question says "at least".

Two Questions of Twenty, and What That Buys

The section gives 20 questions and the paper gives 180 minutes for 100, so you have about 108 seconds a question — generous by CAT standards and the reason CMAT's counting questions are readable rather than tricky. At +4 and −1, a question you can settle in a minute is worth attempting on a two-way guess and worth skipping on a four-way one: a blind guess among four options returns (1/4 × 4) − (3/4 × 1) = +0.25, barely positive, while eliminating two options first takes it to (1/2 × 4) − (1/2 × 1) = +1.5. Counting questions are unusually easy to half-solve, which is what makes them worth the twenty minutes of preparation this chapter asks and no more.

🎯 The one question that comes before every other

Does the order matter? A handshake between A and B is the same handshake as between B and A, so handshakes are combinations. A president and a secretary chosen from the same two people are two different outcomes, so posts are permutations. Ask it before writing anything, because every wrong option in a CMAT counting question is the answer to the same question with the other reading — nCr where nPr was wanted, or nPr where nCr was.

Counting: Five Formulas and When Each Applies

Everything in this block descends from the fundamental counting principle: if one choice can be made in m ways and a second, independently, in n ways, the pair can be made in m × n ways. Permutations and combinations are that principle with the bookkeeping done for you. nPr = n!/(n−r)! counts ordered selections; nCr = n!/(r!(n−r)!) counts unordered ones, and the r! in the denominator is exactly the number of orders you are choosing to ignore. Note also that 0! = 1 and nC0 = nCn = 1, which are the values that make an "at least" calculation come out right.

The five counting situations CMAT actually sets
Situation	Formula	Worked instance
Arrange all n distinct things in a row	`n!`	Five books on a shelf: 5! = 120.
Arrange n things where some repeat	`n!` divided by the factorial of each repeat count	LEVEL has two L and two E: 5!/(2!·2!) = 30.
Arrange n things in a circle	`(n−1)!`	Six guests round a table: 5! = 120, because rotations are the same seating.
Choose r of n, order irrelevant	`nCr`	A committee of 3 from 8: 8C3 = 56.
Choose r of n and assign distinct posts	`nPr`	President and secretary from 8: 8P2 = 56 as well — a coincidence at these numbers, and the reason to derive rather than recognise.

🧠 "At least" means compute the complement

A committee of 4 from 5 men and 4 women with at least two women can be counted directly as three cases — 2W2M, 3W1M, 4W — giving 6×10 + 4×5 + 1 = 81. It can also be counted as everything minus the bad cases: 9C4 = 126, minus the committees with no woman (5C4 = 5) and exactly one woman (4×10 = 40), which is 126 − 45 = 81. Use whichever list is shorter, and when a question says "at least one", the complement is a single term and always shorter.

Probability: One Definition and Two Rules

For equally likely outcomes, P(event) = favourable outcomes ÷ total outcomes, and almost all the work is counting the two numbers correctly — which is why this section follows the counting one. Then the two rules. Addition, for "or": P(A or B) = P(A) + P(B) − P(A and B), where the subtraction is the overlap and is zero only when the events cannot both happen. Multiplication, for "and" with independent events: P(A and B) = P(A) × P(B). And the complement, P(not A) = 1 − P(A), which converts every "at least one" question into a one-line calculation.

Two dice have 36 outcomes, not 12. A sum of 8 happens in five of them — (2,6), (3,5), (4,4), (5,3), (6,2) — so P = 5/36. Ordered pairs are the sample space, which is why (2,6) and (6,2) are counted separately.
A pack has 52 cards: 4 suits of 13, 26 red and 26 black, 12 face cards, 4 of each rank. P(king or heart) = 4/52 + 13/52 − 1/52 = 4/13, and the subtracted card is the king of hearts.
Mutually exclusive is not the same as independent. Two events that cannot both occur are exclusive, so their overlap is zero; independence means one's occurrence does not change the other's probability. Exclusive events are never independent unless one is impossible.
Drawing with replacement keeps the probabilities constant; drawing without replacement does not. A CMAT question always states which, and the phrase to look for is "one after another, without replacement".

⚠️ Adding probabilities that overlap

P(king) + P(heart) = 17/52 is the wrong answer that appears in almost every version of this question, and it is wrong by exactly one card. Before adding, ask whether both events can happen at once; if they can, subtract the overlap. The same discipline catches the other classic: P(even) + P(greater than 3) on one die is not 1/2 + 1/2, because 4 and 6 are in both.

Key takeaways

This is one or two questions of the twenty in CMAT's quant section, at about 108 seconds a question. Prepare it to answer a clean question in a minute, and no further — the rest of the section is worth more of your time.
Ask "does the order matter?" before writing anything. Handshakes and committees are combinations (nCr); posts, ranks and seat numbers are permutations (nPr). Every wrong option is the other reading of the same question.
Repeats divide: LEVEL is 5!/(2!·2!) = 30. Circles lose one: six round a table is 5! = 120. And "at least" is a signal to count the complement — 1 − P(none) is one term where the direct count is three.
Two dice give 36 ordered outcomes; a pack gives 52 cards with 13 per suit. For "or", subtract the overlap: P(king or heart) = 4/13, not 17/52.

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.

Ten people at a meeting each shake hands with every other person exactly once. How many handshakes take place?
1. 90
2. 45
3. 100
4. 20
Show answer
Answer: B — 45
A handshake is an unordered pair, since A shaking hands with B is the same event as B shaking hands with A, so this is 10C2 = (10 × 9)/2 = 45. The 90 option is 10P2, the ordered count, and it is exactly double — which is what happens when the order question is answered wrongly, and it is the option placed first for that reason. 100 is 10 × 10, which counts each person shaking hands with themselves as well. And 20 is 2 × 10, a guess with no counting model behind it.
In how many distinct ways can the letters of the word LEVEL be arranged?
1. 120
2. 30
3. 60
4. 20
Show answer
Answer: B — 30
LEVEL has five letters with L appearing twice and E appearing twice, so the count is 5!/(2! × 2!) = 120/4 = 30. Dividing by the factorial of each repeat count removes the arrangements that merely swap identical letters and are therefore not distinct. 120 is 5! and is the answer for five different letters, which is the commonest error — treating repeated letters as distinct. 60 divides by only one of the two repeats, 2! instead of 2!×2!, and is the near-miss that catches candidates who spot the two L and not the two E. 20 corresponds to no correct step here.
Six guests are to be seated around a circular table. In how many ways can this be done?
1. 720
2. 120
3. 360
4. 30
Show answer
Answer: B — 120
Circular arrangements of n people number (n−1)!, so six guests give 5! = 120. The reduction happens because a rotation of the whole table produces the same seating — only the relative order matters, and fixing one guest's seat as a reference removes the six rotations that would otherwise be counted separately. 720 is 6! and is the answer for a straight row; it is the standard error and the reason this formula is worth memorising with its justification rather than alone. 360 is 6!/2 and would be right only if reflections were also treated as identical, which is a different question — a necklace rather than a table. 30 corresponds to no step.
Two fair dice are thrown together. What is the probability that the sum of the numbers shown is 8?
1. 1/6
2. 5/36
3. 1/9
4. 1/12
Show answer
Answer: B — 5/36
The sample space is 6 × 6 = 36 ordered pairs. The pairs summing to 8 are (2,6), (3,5), (4,4), (5,3) and (6,2) — five of them, so the probability is 5/36. Counting (2,6) and (6,2) separately is essential, and treating them as one outcome is what produces 1/12: three unordered pairs over 36. The 1/9 option is 4/36 and comes from missing (4,4), the pair that is easy to skip because it has no mirror image. And 1/6 is the probability of a single face on one die, which has nothing to do with a sum.
One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is a king or a heart?
1. 17/52
2. 4/13
3. 1/4
4. 13/52
Show answer
Answer: B — 4/13
There are 4 kings and 13 hearts, and one card is both — the king of hearts — so the favourable count is 4 + 13 − 1 = 16, and 16/52 = 4/13. The subtraction is the addition rule doing its work: the two events are not mutually exclusive, so adding the probabilities counts the shared card twice. 17/52 is exactly that error and is the option most often chosen, because 4 + 13 looks like the whole answer. 1/4 is 13/52, the probability of a heart alone, and 13/52 is the same value written unreduced — both drop the kings entirely, which is what happens when "or" is read as naming only the larger set.
A fair coin is tossed three times. What is the probability of getting at least one head?
1. 3/8
2. 7/8
3. 1/2
4. 1/8
Show answer
Answer: B — 7/8
Use the complement. The only way to fail is three tails, whose probability is (1/2)³ = 1/8, so P(at least one head) = 1 − 1/8 = 7/8. Counting directly would mean adding the cases of exactly one, exactly two and exactly three heads — 3/8 + 3/8 + 1/8 — which gives the same answer with three times the work, and the 3/8 option is what a candidate reports after computing only the first of those cases. 1/8 is the complement itself, offered as the answer to catch a subtraction that was set up and then skipped. And 1/2 is the probability for a single toss.
From 8 candidates, a president and a secretary are to be chosen, and no one may hold both posts. In how many ways can this be done?
1. 28
2. 56
3. 64
4. 16
Show answer
Answer: B — 56
The two posts are distinct, so order matters: 8 ways to pick the president and then 7 for the secretary, giving 8 × 7 = 56, which is 8P2. The 28 option is 8C2, the unordered count, and it is the answer to a different question — a committee of two, where naming who is president would be meaningless. 64 is 8 × 8 and allows the same person to hold both posts, which the question forbids. 16 is 2 × 8 and has no model behind it. Note that 8C3 also equals 56; the coincidence is a reminder to derive the count rather than recognise a familiar number.
A committee of 4 is to be formed from 5 men and 4 women, with at least two women. In how many ways can it be formed?
1. 126
2. 81
3. 60
4. 45
Show answer
Answer: B — 81
Split by the number of women: two women and two men is 4C2 × 5C2 = 6 × 10 = 60; three women and one man is 4C3 × 5C1 = 4 × 5 = 20; four women is 4C4 = 1. The total is 60 + 20 + 1 = 81. The complement route agrees: all committees 9C4 = 126, minus none-women 5C4 = 5 and exactly-one-woman 4C1 × 5C3 = 4 × 10 = 40, so 126 − 45 = 81. The 126 option is the unrestricted count, ignoring the condition. 60 is the first case only and is the most frequent error, because "at least two" is read as "exactly two". And 45 is the count of the excluded committees — the quantity you subtract, offered as the answer.