Multiple choice

A bag contains $n$ white, $n$ black balls. Pair of balls are drawn without replacement until the bag is empty. Then the probability that each pair consists of one white and one black ball is

  1. $\displaystyle \frac{n!}{(2n!)}$
  2. $\displaystyle \frac{(n!)^{2}}{(2n)!}$
  3. $\displaystyle \frac{n!.2^{2}}{(2n)!}$
  4. $\displaystyle \frac{(n!)^{2}2^{n}}{(2n)!}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total ways to arrange 2n balls is (2n)!. Ways to pick pairs such that each has one white and one black: there are n! ways to pair white balls with black balls, and 2^n ways to order the balls within each pair. The formula is (n! * 2^n) / (2n!/2^n) which simplifies to (n!)^2 * 2^n / (2n)!. This matches option D.

AI explanation

The total number of ways to arrange 2n balls sequentially in pairs is (2n)! / (n! * n!) because there are n identical white balls and n identical black balls, but since the individual pairs drawn are unordered, we divide by 2^n, making the effective denominator (2n)! / (n! * n! * 2^n). The number of ways to form n specific pairs each containing one white and one black ball is n! * n!. Dividing the favorable outcomes by the total outcomes gives (n! * n!) / ((2n)! / (n! * n! * 2^n)) = (n!)^2 * 2^n / (2n)!.

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