Skip to content
← All questions

How is the counting formula C42(C43C21+C42)C_4^2(C_4^3C_2^1+C_4^2) for exactly two empty boxes derived?

First, choose 2 empty boxes out of 4, which gives C42C_4^2 ways. Then, distribute the 4 distinct balls into the remaining 2 non-empty boxes such that neither is empty. This splits into two cases: a3+1a 3+1 distribution or a2+2a 2+2 distribution. For 3+13+1, choose 3 balls for one box (C43C_4^3) and assign that group to one of the 2 boxes (C21C_2^1). For 2+22+2, choose 2 balls for one designated box (C42C_4^2). The total favorable outcomes are C42∗(C43∗C21+C42)C_4^2 * (C_4^3*C_2^1 + C_4^2).

Conditions

  • The balls are distinct.
  • The boxes are distinct.
  • Exactly two boxes must be empty, meaning the other two must be non-empty.

Reasoning, step by step

  1. Select 2 empty boxes from 4: C42C_4^2.
  2. Identify the two possible occupancy patterns for the remaining 2 boxes: 3+13+1 and 2+22+2.
  3. Calculate ways for 3+13+1 pattern: Choose 3 balls from 4(C43)4 (C_4^3) and choose which of the 2 occupied boxes gets them (C21C_2^1). The remaining 1 ball goes to the other box.
  4. Calculate ways for 2+22+2 pattern: Choose 2 balls from 4 for one specific box (C42C_4^2). The remaining 2 balls go to the other box.
  5. Sum the patterns and multiply by the box selection: C42∗(C43∗C21+C42)C_4^2 * (C_4^3*C_2^1 + C_4^2).

Example

The video writes P(B)=[C42(C43C21+C42)]/44P(B) = [C_4^2(C_4^3 C_2^1 + C_4^2)] / 4^4.

Common misconceptions

  • Forgetting to multiply by C21C_2^1 in the 3+13+1 case, which assigns the group of 3 to a specific box.
  • Double-counting in the 2+22+2 case by treating the boxes as unlabeled (they are distinct, so choosing 2 balls for Box X is different from choosing them for Box Y).

Watch the explanation

Connected concepts

Explore next

Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.