combinatorics problem If you have $5$ bannanas and $5$ apples and two carrots, and the two carrots are lined up with the $10$ fruits at random, what is the probability there are exactly two apples and any number of carrots between the two carrots. Assume that all items are distinct objects.
Attempt:
So there are $10!$ ways to order the $5$ bannanas and $5$ apples. For each ordering, you can place the carrots in $\binom {11}{2}$ slots. So the denominator is $10!*\binom {11}{2}.$
Now we have to find all the points in the sample space where there exactly two apples, and any number of carrots between the two carrots. This is where I am stuck.

If you want to do this fully, there are $12! = 479001600$ arrangements.  But each of the indistinguishable ones are equally probable so you can cut this down to $\frac{12!}{5! \; 5! \; 2!} = 16632$, which is probably what you are aiming for.  You can go further and decide that you do not care about bananas so cutting down to $\frac{7!}{5! \; 2!} = 21$ equally probable arrangements of apples and carrots.
At this stage you could count by hand.  You want carrots to be in positions $k$ and $k+3$ when looking at the order of seven apples and carrots.  So $1 \le k \le 7-3$ giving four possible pairs of positions for the carrots.  So $\frac{4}{21}$ is the answer. 
