Permutation and Combination in probability question - Choose team members I would like to have your help and explanation on following question.
For an 8-a-side football match, a coach has to choose the team from a squad of 12 boys. Only three of them can play as a goalkeeper and these three cannot play any other position. The other boys can play all the other positions – defense, midfield and forward. If the team should have 1 goal keeper, 3 defenders, 3 midfielders and 1 forward, in how many ways can the coach select the team?
a. 495
b. 108
c. 544320
d. 12580
e. 15120

Initially, I was thinking it can be solved in this way:
we have 3 goal keepers that one of them will be chosen (3 options). Aside from them we will have 9 left who can play in different positions, and there is no preference over these 9 so we can choose randomly. Then, we have to use combination and each time we choose for a position (Defense, Mid, Forward) we have to decrease from total amount, which will end like the following.
$3+{9 \choose 3} +{6 \choose 3}+{3 \choose 1}$ = 3 + 84 + 20 + 6 = 113
But result is not in given choices.
Could you please help me where I'm going wrong?

you have to multiply 3*84*20*3 , as for each selection of goal keeper there will be C(9,3) to pick the the defender and so on.
also C(3,1) = 3
