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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.