Of the following, which is the best approximation to the probability that a permutation of size N has exactly 2 cycles?

The probability that a permutation of size N has exactly 2 cycles is approximately 1/N.

Here's a step-by-step explanation:

1. A permutation of size N can be represented as a product of cycles.

2. The number of permutations of size N is N factorial (N!).

3. The number of permutations with exactly 2 cycles is given by the Stirling numbers of the first kind, denoted s(N,2).

4. The Stirling numbers of the first kind count the number of permutations of N elements with a given number of cycles.

5. For large N, the Stirling numbers of the first kind can be approximated as s(N,2) ≈ N!/2.

6. Therefore, the probability that a permutation of size N has exactly 2 cycles is s(N,2)/N! ≈ (N!/2) / N! = 1/2N.

7. However, this is an overestimate because it counts permutations with more than 2 cycles multiple times.

8. A better approximation is 1/N, which is obtained by dividing the number of permutations with exactly 2 cycles by the total number of permutations.

So, the best approximation to the probability that a permutation of size N has exactly 2 cycles is 1/N.