Probability that the first and fourth balls are red when drawing 4 balls.

Question

Sorry, couldn't fit the entire question into the title.

Question:

A box contains 15 identical balls except that 10 are red and 5 are
black. Four balls are drawn successively and without replacement.
Calculate the probability that the first and fourth balls are red.

My attempt:

Probability =
$$1*2C0 + 1*2C1 + 1*2C2 \over 4C0 + 4C1 + 4C2 + 4C3 + 4C4 $$

My idea is that no. of ways to make first and fourth balls = 1, and we have 2 balls left which
can either have red or black colors.

However, my textbook answer was:

$$10P2*13P2\over15P4$$

Which I don't get at all; why would you use permutations when you have identical balls? Wouldn't that mess things up?

Thanks in advance.

Answer 1

Note that the probability of first ball and fourth ball being red is same as probability of first and second ball being red.

So desired probability $ = \displaystyle \frac{10}{15} \cdot \frac{9}{14} = \frac{3}{7}$