COMBINATORICS INTERVIEW QUESTIONS

Most probability mistakes in interviews are really counting mistakes. Get the count right and the probability usually follows in one line.

These are original representative questions, not real or leaked firm questions.

One question to ask first: does order matter, and can things repeat? Most counting mistakes come from getting one of those two wrong.

Arrangements

1. How many distinct arrangements of the letters in LEVEL are there?
   Show worked solution

   Five letters, with L twice and E twice: 5! / (2! · 2!) = 30.

2. Six people sit at a round table. What is the probability that two particular people sit next to each other?
   Show worked solution

   Seat the first person anywhere. The second person has five possible seats and two of them are next to the first, so the answer is 2/5. Counting every arrangement gives the same result more slowly: 2 · 4! / 5! = 48/120. Fixing one person first is the shortcut.

Choosing groups

3. From 8 people, form a committee of 3 with one chair. How many ways?
   Show worked solution

   Pick the committee and then the chair: C(8,3) · 3 = 168. Or pick the chair and then two others: 8 · C(7,2) = 168. When two methods agree, say so. It is the quickest check you can show.

4. How many five-card poker hands are a full house?
   Show worked solution

   Pick the rank for the three of a kind, then its suits, then the rank for the pair, then its suits: 13 · C(4,3) · 12 · C(4,2) = 13 · 4 · 12 · 6 = 3,744. Use 12 for the second rank, not 13, because the pair can't share the first rank.

Stars and bars

5. In how many ways can 10 identical sweets be shared among 4 children?
   Show worked solution

   Lay out 10 stars and 3 bars that split them into four groups: C(13, 3) = 286.

6. Same question, but each child gets at least one.
   Show worked solution

   Give each child one sweet first, then share the remaining 6 freely: C(9, 3) = 84.

7. In how many ways can three dice show a total of 10?
   Show worked solution

   Let each die be 1 + x with 0 ≤ x ≤ 5. The x values then add up to 7. Without the upper limit there are C(9, 2) = 36 ways. Now remove the cases where some x is 6 or more: set that x to 6 plus a remainder, and the remaining 1 is shared among three dice in C(3, 2) = 3 ways. That happens for each of the 3 dice, so 36 − 9 = 27. The probability is 27/216 = 1/8.

Lattice paths

8. Moving only right or up, how many paths go from (0,0) to (4,3)? How many pass through (2,1)?
   Show worked solution

   A path is 7 moves, 3 of them up: C(7,3) = 35. To pass through (2,1), multiply the two legs: C(3,1) · C(4,2) = 3 · 6 = 18.

Overlaps and derangements

9. How many integers from 1 to 100 are divisible by 2 or 3?
   Show worked solution

   Add the two counts, then subtract the numbers counted twice: 50 + 33 − 16 = 67. The 16 is the multiples of 6.

10. Four letters go into four addressed envelopes at random. What is the probability that no letter is in its own envelope?
    Show worked solution

    Count the derangements of 4. You can use inclusion-exclusion, 24 − 24 + 12 − 4 + 1 = 9, or the recurrence D(n) = (n − 1)(D(n−1) + D(n−2)), which gives 3 · (2 + 1) = 9. So the probability is 9/24 = 3/8. For large n it approaches 1/e ≈ 0.368, and 0.375 is already close.

Checking your count

When you're unsure, shrink the problem until you can list every case. Two dice instead of three, or a 2×1 grid instead of 4×3. A mismatch exposes a mistake early. A match is encouraging, but you still need to explain why the formula counts every case exactly once.

Once counting is quick, the probability questions and probability shortcuts will feel much easier.

Small cases are a check, not a proof

The dice are distinguishable, every arrangement is equally likely, and children are distinct even though the sweets are identical. Checking a smaller instance can catch a bad formula, but a matching small case does not prove the general result. State why the objects you count are distinct and why no outcome is missing or counted twice.