Permutations, Combinations and Counting Previous Year Questions (PYQs) for CAT: 8+ Solved Questions with Step-by-Step Solutions

Solve 8+ Permutations, Combinations and Counting previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Permutations, Combinations and Counting

Modern Maths Journey

Permutations, Combinations and Counting

In this chapter, CAT usually checks whether you can convert a word problem into a clean count. The full chapter has 11 own-course PYQs out of 1002 total course questions.

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t1. Counting Arrangements and Product Rule

Learn the base engine: multiply choices across stages, add mutually exclusive routes.

2 PYQs
support-core

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t2. Digit Formation and Non-Repeating Numbers

Count numbers under digit restrictions, zero restrictions, and non-repetition.

6 PYQs
highest weight

🎈

t3. Distribution of Identical Objects

Distribute identical items under lower-bound and parity conditions.

2 PYQs
moderate

✅

t4. Selection with Restrictions

Handle must-include, must-exclude, and cannot-appear-together conditions.

1 PYQ
low-repeat

By the end: you should be able to look at a counting problem and immediately decide: multiply choices, add cases, or break the problem into stages.

Topic Hero: Counting Arrangements and Product Rule

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Selected Topic

Counting Arrangements and Product Rule

The main skill is simple: do not list outcomes; instead, break the action into stages and count choices at each stage.

Stage 1
choices

Stage 2
choices

Stage 3
choices

CAT relevance

This topic has 2 direct PYQs. Its own frequency is moderate, but it is the base logic behind many counting questions.

Permutations, Combinations and Counting: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

The number of all positive integers up to 500 with non-repeating digits is

Correct Answer:

126

Step-by-Step Solution

Pattern: "sum of all three altitudes" of a triangle. Recognise this by the phrase itself — whenever a question asks for altitudes to all three sides, the fastest route is: find the area once, then use $Area = \frac{1}{2} \times base \times height$ three times, once per side.

Why this method works: area is a single fixed number for the triangle. Every side, when used as the base, must pair with its own altitude to reproduce that same area. So once we know the area, each altitude is just Area divided by (half the corresponding side).

Step 1 — Find the altitude to BC.

Since $A B = A C = 50$ , triangle ABC is isosceles, so the altitude from A lands exactly on the midpoint M of BC.

$B M = \frac{B C}{2} = 40$

By Pythagoras in triangle ABM:

$A M = A B^{2} - B M^{2} = 5 0^{2} - 4 0^{2} = 2500 - 1600 = 900 = 30$

Step 2 — Find the area using this altitude.

$Area = \frac{1}{2} \times B C \times A M = \frac{1}{2} \times 80 \times 30 = 1200$

Step 3 — Use the same area to get the altitude onto AC (call it $h_{B}$ , dropped from B).

$Area = \frac{1}{2} \times A C \times h_{B} ⟹ 1200 = \frac{1}{2} \times 50 \times h_{B} ⟹ h_{B} = 48$

Step 4 — Get the altitude onto AB (call it $h_{C}$ , dropped from C).

Since $A B = A C = 50$ (the triangle is isosceles), this altitude uses the identical base length as Step 3:

$1200 = \frac{1}{2} \times 50 \times h_{C} ⟹ h_{C} = 48$

Step 5 — Add up all three altitudes.

$A M + h_{B} + h_{C} = 30 + 48 + 48 = 126$

Common trap: assuming all three altitudes are equal (that is only true for an equilateral triangle). Here two altitudes are equal (because two sides are equal) but the third (to the unequal side BC) is different.

Final answer: 126

Question 2 · Quantitative Ability NAT

The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Correct Answer:

Step-by-Step Solution

This is a must-include-plus-forbidden-pair selection question. You can recognise it from three signal phrases in the question: "always include 3 and 5" (two elements are compulsory), "7 and 8 are never included together" (one specific pair is banned from co-occurring), and "three or more" (a minimum group size is demanded).

Why this method applies: whenever some elements are locked into every valid group, you only need to decide the fate of the remaining elements. Whenever two elements can never both appear, the cleanest way is to count everything freely first and then subtract the cases where the forbidden pair occurs together. The minimum-size rule is applied last, as a final check on the smallest group.

Step 1 — Lock in the compulsory elements.

Every valid group already contains $3$ and $5$ . So the only real decision is which of the remaining six numbers ${1, 2, 4, 6, 7, 8}$ to add on top of ${3, 5}$ .

Step 2 — Count all ways to add these six numbers, ignoring restrictions for now.

Each of the $6$ numbers is independently either included or excluded, so there are

$2^{6} = 64$

possible ways to add extra numbers, including the case of adding nothing at all.

Step 3 — Remove the subsets that break the "never 7 and 8 together" rule.

If both $7$ and $8$ are forced into the group, the other $4$ numbers ${1, 2, 4, 6}$ can still be freely included or excluded, giving

$2^{4} = 16$

forbidden subsets (subsets where $7$ and $8$ are both present).

So the number of ways to add extras while respecting the 7-8 rule is

$64 - 16 = 48$

Step 4 — Apply the "three or more" condition.

A group's total size is $2 + (number of extra numbers added)$ , since ${3, 5}$ alone already gives size $2$ . For the total size to be $3$ or more, at least one extra number must be added. The case of adding nothing (all six extra numbers excluded) is counted inside the $48$ above but is not a valid group, so it must be removed:

$48 - 1 = 47$

Final answer: $47$ .

Common traps: stopping at Step 3 and reporting $48$ forgets that adding nothing leaves the 2-element group ${3, 5}$ , which violates "three or more". A second trap is misreading "never included together" as "never included at all" — that would wrongly remove every group that contains $7$ or every group that contains $8$ individually, instead of only removing groups where both appear.

Question 3 · Quantitative Ability MCQ

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

A.
1440
B.
1200
C.
1480
D.
1420

Correct Answer:

Step-by-Step Solution

This is a "counting numbers greater than a limit, with digits used at most once" question. The signal is the phrase "at most once" (not "exactly once"), which means the number can have any length up to 6 digits, not just 4 digits. This is different from digit-formation questions where all given digits must be used.

Step 1: Identify possible number lengths.

We have 6 digits available: $0, 1, 2, 3, 4, 5$ . Since digits can repeat at most once (i.e., no repetition, but not all digits need to appear), a valid number can have 1, 2, 3, 4, 5, or 6 digits. But we only care about numbers greater than 2000, so we only need to check 4-digit, 5-digit, and 6-digit numbers (1, 2, 3-digit numbers are always less than 2000).

Step 2: Count valid 4-digit numbers greater than 2000.

A 4-digit number cannot start with 0. To be greater than 2000, the first digit must be $2, 3, 4,$ or $5$ (since first digit $1$ gives numbers between $1000$ – $1999$ , all less than 2000; and repetition of digits is not allowed so 2000 itself, which needs three 0's, cannot form).

For each of these 4 choices of first digit, the remaining 3 positions are filled from the remaining 5 digits (which include 0), arranged without repetition:

$^{5} P_{3} = 5 \times 4 \times 3 = 60$

So 4-digit numbers greater than 2000: $4 \times 60 = 240$ .

Step 3: Count all 5-digit numbers.

Any 5-digit number is automatically greater than 2000 (since the smallest 5-digit number, 10234, is far bigger than 2000).

First digit: cannot be 0, so 5 choices (1 through 5).

Remaining 4 positions: arrange from the remaining 5 digits (including 0):

$^{5} P_{4} = 5 \times 4 \times 3 \times 2 = 120$

Total 5-digit numbers: $5 \times 120 = 600$ .

Step 4: Count all 6-digit numbers.

A 6-digit number uses all 6 digits exactly once. First digit: cannot be 0, so 5 choices. Remaining 5 digits arranged in the remaining 5 positions: $5! = 120$ .

Total 6-digit numbers: $5 \times 120 = 600$ .

Step 5: Add all cases.

$240 + 600 + 600 = 1440$

So the answer is $1440$ .

Common trap: many students only count 4-digit numbers because "2000" looks like a 4-digit boundary, forgetting that "at most once" allows longer numbers too, all of which are automatically greater than 2000.

Question 4 · Quantitative Ability MCQ

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

A.
2222
B.
2442
C.
2592
D.
3333

Correct Answer:

Step-by-Step Solution

This is a "mean of all rearrangements with a repeated digit" question. The signal is the word "mean" combined with rearranging a number that has a repeated digit (here, digit 1 appears twice in 1421). This means the standard "each digit repeats equally" trick from the non-repeated case needs a small adjustment.

Step 1: Count total distinct arrangements.

The digits of 1421 are $1, 4, 2, 1$ — that is, two 1's, one 4, one 2. Since two digits are identical, the total number of distinct 4-digit arrangements is:

$\frac{4 !}{2 !} = \frac{24}{2} = 12$

Step 2: Find how many times each digit appears in a given position.

To find how many arrangements have a specific digit fixed in one position, fix that digit and count distinct arrangements of the remaining 3 digits.

Fix digit 2: remaining digits are ${1, 1, 4}$ (two 1's). Arrangements = $\frac{3 !}{2 !} = 3$ .
Fix digit 4: remaining digits are ${1, 1, 2}$ (two 1's). Arrangements = $\frac{3 !}{2 !} = 3$ .
Fix digit 1 (one of the two 1's): remaining digits are ${1, 2, 4}$ (all distinct, since only one 1 is left). Arrangements = $3! = 6$ .

Check: $3 + 3 + 6 = 12$ , which matches the total number of arrangements — this confirms the counts are correct.

Step 3: Find the sum contributed by one position.

In any one position (say units), digit 1 appears 6 times, digit 2 appears 3 times, digit 4 appears 3 times:

$6 (1) + 3 (2) + 3 (4) = 6 + 6 + 12 = 24$

Step 4: Extend to all four positions.

Since every position behaves identically by symmetry, the total sum over all 12 numbers is:

$24 \times 1111 = 26664$

Step 5: Compute the mean.

$Mean = \frac{26664}{12} = 2222$

So the answer is $2222$ .

Common trap: students often use the non-repeated-digit shortcut directly (treating each of the 4 digits as appearing $3! = 6$ times per position), which is wrong here because digit 1 appears twice in the original number — this uneven repetition means each digit does NOT appear the same number of times per position, and skipping the "fix and count" step leads to an incorrect mean.

Question 5 · Quantitative Ability MCQ

The number of all natural numbers up to 1000 with non-repeating digits is

A.
504
B.
648
C.
738
D.
585

Correct Answer:

Step-by-Step Solution

This is a "counting numbers with non-repeating digits up to a limit" question. The signal words are "up to 1000" combined with "non-repeating digits" — this means you must break the count into cases by number of digits, because the counting rule changes with position and length.

Step 1: Break into cases by digit-length.

Natural numbers from 1 to 1000 fall into 1-digit, 2-digit, 3-digit, and the single 4-digit number 1000.

Step 2: Count 1-digit numbers.

These are 1 through 9. Every single digit is automatically non-repeating. Count = 9.

Step 3: Count 2-digit numbers with distinct digits.

First digit: cannot be 0 (else it wouldn't be a 2-digit number), so 9 choices (1–9).
Second digit: any digit 0–9 except the one already used, so 9 choices.

$9 \times 9 = 81$

Step 4: Count 3-digit numbers with distinct digits.

First digit: 9 choices (1–9, no zero)
Second digit: 9 choices (0–9 except the first digit)
Third digit: 8 choices (0–9 except the two already used)

$9 \times 9 \times 8 = 648$

Step 5: Check 1000 itself.

$1000$ has the digits $1, 0, 0, 0$ — the digit 0 repeats three times, so it does NOT qualify as having non-repeating digits. It is excluded.

Step 6: Add everything up.

$9 + 81 + 648 = 738$

So the answer is $738$ .

Common trap: a very common mistake is to forget that 0 is allowed in the second and later positions (just not the first), which artificially shrinks the count. Another common mistake is forgetting to add the 1-digit and 2-digit counts and reporting only the 3-digit count.

Permutations, Combinations and Counting Previous Year Questions (PYQs) for CAT: 8+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Permutations, Combinations and Counting

Permutations, Combinations and Counting

Topic Hero: Counting Arrangements and Product Rule

Counting Arrangements and Product Rule

Permutations, Combinations and Counting: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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