Doubt on probability

Question

15 students among whom are A and B arranged randomly in a row. Obtain the probability that there will be 3 students between A and B?

Answer 1

Total number of permutation possible = 15! - that's our sample space.

Now for favourable cases, select three which will be present in the middle. This can be done in $13 \choose 3$. Now, we have 11 blocks as the group of AC1C2C3B have to always be together.

Within the block, the children can be arraged in 3! ways.

So answer should be: $13 \choose 3$$\times$$\frac{ 11! \times 3!}{15!}$

Comments

You have not taken 13C3 why..?

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Oh sorry, I missed that while typing, I'll edit it.