Multiple choice

An unbiased die, with faces numbered $1, 2, 3, 4, 5, 6$ is thrown n times and the list of n numbers showing up is noted. Then the probability that, among the numbers $1,2, 3,4, 5,6$ only three numbers appear in this list ?

  1. $\displaystyle \frac{[3^{n}-3(2^{n})+3]}{6^{n}}$
  2. $\displaystyle \frac{^{6} C_{3}[3^{n}-3(2^{n})+3]}{6^{n}}$
  3. $\displaystyle \frac{^{6} C_{3}[3^{n}-3(2^{n})]}{6^{n}}$

  4. None of these
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B Correct answer
Explanation

This is a combinatorics problem involving the inclusion-exclusion principle for surjective mappings. The number of ways to choose 3 numbers out of 6 is 6C3. The number of ways to map n throws to exactly 3 specific numbers is 3^n - 3(2^n) + 3.

AI explanation

The total number of possible lists is 6^n. We choose the 3 specific numbers that will appear in 6C3 ways. By the principle of inclusion-exclusion, the number of ways to form a list using exactly these 3 numbers (with all 3 appearing at least once) is 3^n - 3(2^n) + 3. Multiplying the choices and dividing by the total outcomes gives the probability as 6C3 * [3^n - 3(2^n) + 3] / 6^n.

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