Find the probability that the two Tan brothers are selected for the soccer team A soccer team is made of eleven players. One goal keeper, four defenders, four midfielders and two attackers. Plus Ultra FC has twenty players in its squad. There are two goal keepers, six defenders, seven midfielders and five attackers in its squad. The Tan brothers are two players in the squad. One is attacker and the other is a midfielder. Plus Ultra FC has to form up a team for an upcoming friendly match, find the probability that. 
(i) both the Tan brothers are selected for the team, and 
(ii) at most one of the Tan brothers and the older goal keeper are selected. 
My attempt: 
(i) Total number of ways to form a team is $2 \choose 1$ $6\choose4$ $7\choose4$ $5\choose2$. If both Tan brothers are selected, then we have $2\choose1$ $6\choose4$ $6\choose3$ $4\choose1$. Then we use the latter divided by the former. Is this correct?
(ii) No idea how to solve it. Any hint is appreciated. 
 A: For (i) You select 3 midfielders out of 6 ( because one of the brother has occupied one place) and 1 attacker out of 4 (same reason). Thus $C(6,3)\times C(4,1)\times C(2,1)\times C(6,4)$. this includes your goalkeeper and defenders too .
For (ii) You take the probability that At most one brother is selected, this means (Total-Both are selected) Although, we will strike that goal keeper selection because one old guy is already selected in the team.
Which makes it $C(7,4)\times C(5,2)\times C(6,4)$ - $C(6,3)\times C(4,1)\times C(6,4)$
Also, personal request. It's football, not soccer. Thank you
A: (i) The attacker has $2/5$ chances being chosen, the midfielder $4/7$. These are independent, so the chances for both to occur is just $\frac{2}{5} \cdot \frac{4}{7}=\frac{8}{35}\approx22.86\%$
(ii) $1/2$ for the older goalkeeper. $1-\frac{8}{35}$ for not-both being chosen. Again you can multiply the two: $\frac{27}{35} \cdot \frac{1}{2}=\frac{27}{70}\approx 38.57\%$
BTW: your solution for (i) is not wrong, it is just overkill. You get the same result
