# 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.