The number $X$ of defective phones selected has a hypergeometric distribution. In R, the direct computation for $P(X = 2)$ is as follows:

dhyper(2, 2,23, 10)
[1] 0.15

The computation amounts to $$P(X = 2) = \frac{{2\choose 2}{23\choose 8}}{{25 \choose 10}} = 0.15,$$ where ${2 \choose 2}= 1$ may be omitted.

choose(2,2)*choose(23,8)/choose(25,10)
[1] 0.15

If you ask for the probability of getting exactly one defective phone then it's 0.5. Solution using binomial coefficients will have ${2 \choose 1} = 2$ in the numerator, which can't be omitted: $$ P(X = 1) = \frac{{2\choose 1}{23\choose 9}}{{25 \choose 10}} = 0.5. $$

dhyper(1,2,23,10)
[1] 0.5
choose(2,1)*choose(23,9)/choose(25,10)
[1] 0.5