Combinations

[riteshb]

Katie has 9 employees that she must assign to 3 different
projects. If 3 employees are assigned to each project & no one
is assigned to multiple projects, how many different combinations
of project assignments are possible?

[n_adroja]

Hi Ritesh,

is it84*20 = 1680?
adroja

Katie has 9 employees that she must assign to 3 different
projects. If 3 employees are assigned to each project & no one
is assigned to multiple projects, how many different combinations
of project assignments are possible?

[riteshb]

That's right, can you explain this one? How you got 84 and 20?

[nikishin]

9!/[3!(9!-3!)]=84
6!/[3!(6!-3!)]=20

[alakshma]

From 9 employees you have to choose 3 for the first group. This can be done in 9C3=84 ways.

Now, from 6 employees you have to choose 3 employees and this can be done in 6C3=20 ways.

From 3 empolyees choosing 3 is 1 way.

Hence 84*20*1=1680

That's right, can you explain this one? How you got 84 and 20?