Probability of combinations so that only $2$ of the $3$ types remains. $15$ telephones have just been received at an authorized service center. $5$ of  these telephones are cellular, $5$ are cordless, and the other $5$ are corded phones. Suppose these components are randomly allocated the number $1,2,3, \dots$.
What is probability that after servicing $10$ of these phones, phones of only $2$ of the $3$ types remain to be serviced?

To start, lets fix a telephone type, solve the problem for that type and then multiply with $3$ the obtained probability to obtain the correct answer (due to symmetry). We need to service all $5$ phones of this type and at least $1$ phone of each of the other $2$ types (in order to exhaust exactly this type and not $2$ types). The probability for this is (hypergeometric) $$\dfrac{\dbinom{5}{5}\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}=\dfrac{\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}$$ Since there are $3$ types it remains to multiply this probability with $3$ to obtain the answer $$3\cdot\dfrac{\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}=3\cdot\frac{2\dbinom{5}{1}^2+2\dbinom{5}{2}^2}{\dbinom{15}{10}}=\frac{750}{3003}\approx 0,25$$
