24 questions · CET 2025–2026 · Easy 12 · Medium 11 · Hard 1. Newest paper first. Full papers
In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty ?
Answer: D official key
Approach: Identical objects into distinct boxes, none empty → (n − 1)C(r − 1).
Solution:
n = 11 toys, r = 3 boxes
Ways = (11 − 1)C(3 − 1) = 10C2 = (10 × 9)/2 = 45
Cetking shortcut: Put 1 toy in each box first; distribute the remaining 8 freely: (8 + 2)C2 = 10C2 = 45.
Answer: D
7 digit number comprises of only 2's and 3's. How many of these 7 digit numbers shall be multiples of 12 ?
Official CET key: E · Cetking's answer: C — see why below
Approach: Divisible by 12 ⇔ divisible by 4 and by 3.
Solution:
Divisible by 4: last two digits from {22, 23, 32, 33} → only 32 works.
First 5 digits: let k of them be 2 and (5 − k) be 3.
Digit sum = 2k + 3(5 − k) + 3 + 2 = 20 − k
Divisible by 3 → 20 − k ∈ {18, 15} → k = 2 or k = 5
k = 2: 5C2 = 10 numbers; k = 5: 5C5 = 1 number
Total = 10 + 1 = 11
Cetking shortcut: Fix the ending first (32), then the count of 2's in the rest must make the sum a multiple of 3.
Answer: E
Arnav wants to print all numbers from 100 to 10000 on a blank paper, how many times would the digit 3 be printed by Arnav on the paper ?
Official CET key: C · Cetking's answer: B — see why below
Approach: Count the digit 3 in 0000–9999, then remove 0–99.
Solution:
In 0000–9999 each of the 4 places shows 3 exactly 1000 times → 4 × 1000 = 4000.
In 0–99: units place 10 times (3, 13, …, 93) + tens place 10 times (30–39) = 20.
10000 has no 3.
From 100 to 10000 = 4000 − 20 = 3980
Cetking shortcut: Each digit appears equally in every place: places × 10^(places−1) = 4 × 1000, then subtract the excluded range.
Answer: C
How many numbers of up to 5 digits can be created by using the digits 1, 2, 3 and 5 each, at least once, such that the numbers are always a multiple of 15 ?
Official CET key: C · Cetking's answer: D — see why below
Approach: Multiple of 15 ⇔ ends in 5 (digit 0 unavailable) and digit sum divisible by 3.
Solution:
All of 1, 2, 3, 5 must appear → at least 4 digits.
4 digits: 1 + 2 + 3 + 5 = 11, not divisible by 3 → none.
5 digits: one extra digit x from {1, 2, 3, 5}; sum = 11 + x must be divisible by 3 → x = 1 (sum 12).
Digits 1, 1, 2, 3, 5 with 5 fixed at the end.
Arrange 1, 1, 2, 3 in the first four places = 4!/2! = 12
Cetking shortcut: Check the 4-digit sum first (11) — the only fix to reach a multiple of 3 within the allowed digits is one more 1.
Answer: C
In how many ways can the alphabets of the word ATTITUDE be rearranged in such a way that no two T's are adjacent to each other ?
Answer: A official key
Approach: Gap method — arrange the other letters, then place the T's in separate gaps.
Solution:
ATTITUDE: 8 letters; T × 3, and A, I, U, D, E all distinct.
Arrange A, I, U, D, E: 5! = 120 ways.
They create 6 gaps; choose 3 for the identical T's: 6C3 = 20.
Total = 120 × 20 = 2400
Answer: A
At the end of a business conference all the 16 people present, shake hands with each other only once. How many handshakes were there altogether?
Answer: B official key
Approach: Each handshake is a pair of people: ⁿC₂.
Solution: Number of handshakes = ¹⁶C₂ = 16 × 15 / 2 = 120.
Cetking shortcut: n(n − 1)/2 – for 16 people, 8 × 15 = 120.
Answer: B
Answer: D official key
Approach: Divisible by 5 ⇒ units digit is 0 or 5; only 5 is available.
Solution:
Units digit fixed = 5 (1 way).
Hundreds digit: any of the remaining 5 digits.
Tens digit: any of the remaining 4 digits.
Total = 5 × 4 × 1 = 20.
Why the other options are wrong:
A (5) – fills only one free place.
C (15), E (25) – allow repetition or wrong count.
B (10) – uses combinations (5C2).
Cetking shortcut: Fix the restricted place first, then fill the rest: 5P2 = 20.
Answer: D
At Rajendra’s Birthday party, there are exactly 57 persons including Rajendra. All of them shake hand with each other. Find the total number of handshakes ?
Answer: A official key
Approach: Each handshake is a pair of people → ⁿC₂.
Solution:
Handshakes = 57 × 56 ÷ 2
= 57 × 28
= 1596.
Answer: A
Answer: D official key
Approach: Women together → treat them as one block.
Solution:
Units: 5 men + 1 women-block = 6 → 6! = 720 ways.
Women inside the block: 4! = 24 ways.
Total = 720 × 24 = 17,280.
Why the other options are wrong:
Answer: D
Answer: B official key
Approach: Place vowels in their allowed positions, then consonants in the rest.
Solution:
EDUCATION: 9 distinct letters — vowels E, U, A, I, O (5), consonants D, C, T, N (4).
Odd positions 1, 3, 5, 7, 9 = 5 places → vowels: 5! = 120.
Remaining 4 even places → consonants: 4! = 24.
Total = 120 × 24 = 2880.
Why the other options are wrong:
Cetking shortcut: 5 vowels exactly fill 5 odd slots → answer is simply 5! × 4!.
Answer: B
DIRECTION for the question: Solve the following question and mark the best possible option.
In how many different ways can the letters of the word OPTICAL can be arranged in such a way that the vowels always come together?
Answer: B official key
Approach: Block method.
Solution: OPTICAL has 7 distinct letters: vowels O, I, A (3); consonants P, T, C, L (4).
Vowel block + 4 consonants = 5 units → 5! = 120 ways.
Vowels inside the block: 3! = 6 ways.
Total = 120 × 6 = 720.
Why the other options are wrong:
A (120): forgets internal arrangement.
E (6): only the internal arrangement.
C, D: wrong block counts.
Answer: B
Answer: D
Approach: Vowels together: tie the vowels into one block, divide by repeats.
Solution:
ESPECIALLY = E, S, P, E, C, I, A, L, L, Y (10 letters).
Vowels: E, E, I, A (4); consonants: S, P, C, L, L, Y (6, L repeated).
Units to arrange = 1 vowel block + 6 consonants = 7 ⇒ 7!/2! = 2520 (2! for the two Ls).
Inside the block: 4!/2! = 12 (2! for the two Es).
Total = 2520 × 12 = 30240.
Why the other options are wrong:
B 25200, C 10080: forget one of the repeat divisions or the block's internal arrangement.
A, E: no standard method gives them.
Cetking shortcut: Answer must be divisible by 7!/2! = 2520 — only 25200, 10080 and 30240 are; 2520 × 12 = 30240.
Answer: D
Three Englishmen and three Frenchmen work for the same company Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?
Answer: C
Approach: Treat it as a gossip problem with a restriction: only the bilingual Englishman (E3) can talk to both groups. Use the 4-call core trick so that only 8 calls are needed.
Solution:
Englishmen E1, E2, E3 (E3 knows French); Frenchmen F1, F2, F3.
1. E1–E3 (English) and 2. E2–E3 (English): E3 now knows E1, E2, E3.
3. F1–F2 (French): they know F1, F2.
4. E3–F3 (French): both know E1, E2, E3, F3.
5. F1–E3 (French): both now know all 6 secrets.
6. F2–F3 (French): both now know all 6 secrets.
7. E3–E1 and 8. E3–E2 (English): E1 and E2 now know all 6.
Every call uses a language both speakers know, and everyone knows all six secrets after 8 calls.
8 is also the minimum: for 6 people at least 2n − 4 = 8 calls are needed even with no language restriction.
Why the other options are wrong:
A (10) and B (9): possible but not minimal — e.g. 'gather at E3 then spread' takes 9.
D (7), E (6): fewer than the 2n − 4 = 8 lower bound, so impossible.
Answer: C
In a chess competition involving some boys and girls of a school: every student had to play exactly one game with every other student It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is
Answer: C
Approach: Games among n players = nC2; mixed games = boys × girls.
Solution:
Girls: G(G − 1)/2 = 45 → G(G − 1) = 90 = 10 × 9 → G = 10
Boys: B(B − 1)/2 = 190 → B(B − 1) = 380 = 20 × 19 → B = 20
Boy-girl games = 20 × 10 = 200
Cetking shortcut: Recognise 45 = 10C2 and 190 = 20C2 instantly.
Answer: C
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: C
Approach: Combinations — case-wise addition.
Solution:
3 men + 2 women: 7C3 × 6C2 = 35 × 15 = 525.
4 men + 1 woman: 7C4 × 6C1 = 35 × 6 = 210.
5 men: 7C5 = 21.
Total = 525 + 210 + 21 = 756.
Why the other options are wrong:
A, B, D, E: miss a case or miscount 6C2 = 15.
Answer: C
In how many ways can the letters of the word 'LEADER' be arranged?
Answer: A
Approach: Permutations with identical letters.
Solution:
LEADER: 6 letters with E repeated twice (L, A, D, R once).
Arrangements = 6!/2! = 720/2 = 360.
Why the other options are wrong:
B, C, D, E: not 720/2.
Answer: A
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
Answer: C
Approach: nPr for ordered selections.
Solution:
LOGARITHMS has 10 distinct letters (L, O, G, A, R, I, T, H, M, S).
4-letter words without repetition = 10P4 = 10 × 9 × 8 × 7 = 5040.
Why the other options are wrong:
A, B, D, E: not 10 × 9 × 8 × 7.
Answer: C
P, Q, R and S are four towns. One can travel between P and Q along 3 direct paths, between Q and S along 4 direct paths, and between P and R along 4 direct paths. There is no direct path between P and S, while there are few direct paths between Q and R, and between R and S. One can travel from P to S either via Q, or via R, or via Q followed by R, respectively, in exactly 62 possible ways. One can also travel from Q to R either directly, or via P, or via S, in exactly 27 possible ways. Then, the number of direct paths between Q and R is.
Answer: E
Approach: Multiplication principle for routes; set up equations in the unknown path counts.
Solution:
Let Q–R = m direct paths and R–S = n direct paths; P–Q = 3, Q–S = 4, P–R = 4, P–S = 0.
P to S: P→Q→S = 3×4 = 12; P→R→S = 4n; P→Q→R→S = 3mn.
12 + 4n + 3mn = 62 → 4n + 3mn = 50 … (1)
Q to R: direct = m; Q→P→R = 3×4 = 12; Q→S→R = 4n.
m + 12 + 4n = 27 → m + 4n = 15 … (2)
(1) − (2): 3mn − m = 35 → m(3n − 1) = 35 = 7 × 5.
3n − 1 = 5 gives n = 2, m = 7; check (2): 7 + 8 = 15 ✓.
Cetking shortcut: From (2), m = 15 − 4n; try n = 1, 2, 3 → only n = 2 (m = 7) satisfies (1): 8 + 42 = 50.
Answer: E
In how many ways can the letters of the word MATHEMATICS be arranged so that all the vowels always come together?
Answer: D
Approach: Glue the vowels into one block; divide by factorials of repeated letters.
Solution:
MATHEMATICS: 11 letters; vowels A, A, E, I; consonants M, M, T, T, H, C, S.
Units = 7 consonants + 1 vowel block = 8 → 8!/(2! × 2!) = 40320/4 = 10,080 (M and T repeat).
Inside the block: 4!/2! = 12 (A repeats).
Total = 10,080 × 12 = 1,20,960.
Why the other options are wrong: 4989600 is all arrangements with no condition (11!/(2!2!2!)); 120800 and 20660 are calculation slips; 11 is the letter count.
Answer: D
Answer: A
Approach: Count blocks of consecutive boxes (a block of length k in n boxes can start at n − k + 1 places).
Solution:
The green balls form one unbroken block of length k (k = 1 to 6); every other box is red.
k = 1: 6 ways; k = 2: 5; k = 3: 4; k = 4: 3; k = 5: 2; k = 6: 1.
Total = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Cetking shortcut: Number of consecutive blocks in n positions = n(n + 1)/2 = 6 × 7/2 = 21.
Answer: A
Answer: A
Approach: Divisible by 6 = even last digit + digit sum divisible by 3. Pick the digit set first, then arrange.
Solution:
Sum of 0, 2, 3, 4, 6 = 15. Dropping one digit must leave a multiple of 3 → drop 0, 3 or 6.
{2, 3, 4, 6}: last digit 2/4/6 (3 ways) × 3! = 18.
{0, 2, 4, 6}: all even; arrangements without leading 0 = 4! − 3! = 18.
{0, 2, 3, 4}: ends in 0 → 3! = 6; ends in 2 → first digit from {3, 4} (2) × 2 = 4; ends in 4 → 4. Total 14.
Total = 18 + 18 + 14 = 50.
Cetking shortcut: Check the total digit sum first (15) — only drop digits that are multiples of 3.
Answer: A
Answer: C
Approach: By symmetry each position holds each digit equally often; average digit × 1111.
Solution:
Digits 1, 1, 2, 4 → distinct arrangements = 4!/2! = 12.
In each place, the average digit = (1 + 1 + 2 + 4)/4 = 2.
Average number = 2 × 1000 + 2 × 100 + 2 × 10 + 2 = 2 × 1111 = 2222.
Cetking shortcut: Mean of all permutations = (digit sum / number of digits) × 111…1.
Answer: C
Answer: D
Approach: Identical items with minimums: give the minimums first, then distribute the rest freely (stars and bars).
Solution: Minimums: Aai 1 + Bai 2 + Cai 3 = 6 pens.
Remaining = 8 − 6 = 2 pens, distributed freely among 3 people.
Non-negative solutions of x + y + z = 2: C(2 + 3 − 1, 3 − 1) = C(4, 2) = 6.
(List: 200, 020, 002, 110, 101, 011.)
Cetking shortcut: Ways to give r identical items to n people (zero allowed) = C(r + n − 1, n − 1).
Answer: D
Answer: D
Approach: Fundamental counting principle, most-restricted task first.
Solution:
Task 2 → person 3 or person 4: 2 ways
Task 1 → not person 1, not person 2, not the person who took Task 2: 6 − 3 = 3 ways
Remaining 4 tasks → remaining 4 persons: 4! = 24 ways
Total = 2 × 3 × 24 = 144
Why the other options are wrong:
B (192): forgets that Task 2's person is used up (2 × 4 × 24).
A (72): counts only one choice for Task 2.
C (360), E (256): no valid counting gives these.
Answer: D
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