Simple basic math problem

[jas_fluidm]

a manager is forming a 6-person team to work on a certain project . from the 11 candidates available for the team , the manager has already chosen 3 to be on the team , in selecting the other 3 team members , how many different combinations of 3 of the remaining candidates does the manager have to choose from ?

answer:
There are only 8 people left to choose from, so 8C3 = 8! / (3! * 5!) = (8*7*6*5*4*3*2*1)/(3*2*1*5*4*3*2*1) = (8*7*6)/(3*2*1) = 56

[arefin]

Thanks

[Sima]

Truly basic...

[tasnuva]

Thanks for sharing..