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CAT QuanthardModern Maths

Modern Maths: Counting, Probability, Progressions and Sets

Updated 22 October 2026

Five questions on permutations and combinations, probability, progressions and set theory.

Question 1 · Multiple choice (+3 / −1)

In how many ways can the letters of the word CATALYST be arranged so that the two A's are not next to each other?

Show the solution

Answer: C. 7,560

  1. CATALYST has 8 letters with A and T each appearing twice.
  2. Total arrangements = 8!/(2! × 2!) = 40,320/4 = 10,080.
  3. Arrangements with the A's together: treat AA as one block, giving 7 units with T repeated: 7!/2! = 2,520.
  4. Required = 10,080 − 2,520 = 7,560.
  5. Alternative: arrange the other 6 letters in 6!/2! = 360 ways and choose 2 of the 7 gaps for the A's: 360 × 21 = 7,560.
Question 2 · Type in the answer (+3 / 0)

A fair coin is tossed 5 times. What is the probability that no two heads occur on consecutive tosses? Enter the answer as a decimal.

Show the solution

Answer: 0.40625

  1. Let aₙ be the number of sequences of n tosses with no two consecutive heads. A valid sequence ends in T (any valid sequence of n − 1 before it) or in TH (any valid sequence of n − 2 before it), so aₙ = aₙ₋₁ + aₙ₋₂.
  2. a₁ = 2 (H, T) and a₂ = 3 (HT, TH, TT).
  3. a₃ = 5, a₄ = 8, a₅ = 13.
  4. Probability = 13/32 = 0.40625.
Question 3 · Multiple choice (+3 / −1)

What is the sum of all three-digit numbers that leave a remainder of 2 when divided by 7?

Show the solution

Answer: C. 70,692

  1. The smallest such number is 100 (= 7 × 14 + 2) and the largest is 996 (= 7 × 142 + 2).
  2. They form an AP with common difference 7. Number of terms = (996 − 100)/7 + 1 = 128 + 1 = 129.
  3. Sum = (number of terms/2) × (first + last) = 129/2 × 1,096 = 129 × 548 = 70,692.
  4. Forgetting the +1 in the term count gives 128 × 548 = 70,144; summing the three-digit multiples of 7 (105 to 994) instead gives 70,336.
Question 4 · Type in the answer (+3 / 0)

In a group of 120 students, 70 study Economics, 60 study Mathematics and 50 study Statistics. 30 study both Economics and Mathematics, 25 study both Mathematics and Statistics, 28 study both Economics and Statistics, and 10 study none of the three. How many students study exactly one of the three subjects? Enter the number of students.

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Answer: 53

  1. Students studying at least one subject = 120 − 10 = 110.
  2. By inclusion–exclusion: 110 = 70 + 60 + 50 − 30 − 25 − 28 + x, where x studies all three. So 110 = 97 + x and x = 13.
  3. Exactly two: (30 − 13) + (25 − 13) + (28 − 13) = 17 + 12 + 15 = 44.
  4. Exactly one = 110 − 44 − 13 = 53.
  5. Check by region: Economics only 25, Mathematics only 18, Statistics only 10; 25 + 18 + 10 = 53.
Question 5 · Multiple choice (+3 / −1)

In how many ways can 10 identical chocolates be distributed among 4 children so that each child gets at least 1 and no child gets more than 4?

Show the solution

Answer: B. 44

  1. Give each child 1 chocolate first. The remaining 6 are shared so that each child gets 0 to 3 more.
  2. Without the upper limit: C(6 + 3, 3) = C(9, 3) = 84 ways.
  3. Subtract cases where some child gets 4 or more extra: fix that child at 4 extra, leaving 2 to share among 4: C(5, 3) = 10 ways, for each of 4 children: 40.
  4. Two children cannot both get 4 extra (that needs 8 > 6), so nothing to add back.
  5. Total = 84 − 40 = 44.

Full worked solution

  1. Arrangements with repeated letters: n!/(p! q! …). For 'not together', subtract the 'together' count or use the gap method.
  2. Binary strings with no two consecutive 1s follow the Fibonacci pattern aₙ = aₙ₋₁ + aₙ₋₂.
  3. AP: number of terms = (last − first)/d + 1; sum = n/2 × (first + last).
  4. Three sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |C ∩ A| + |A ∩ B ∩ C|.
  5. Identical items into r groups, sum n with each ≥ 0: C(n + r − 1, r − 1); handle upper limits by inclusion–exclusion.