1. Good post? |

## Combinations

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?

2. Good post? |

## Re: Combinations

Hi Ritesh,

is it84*20 = 1680?
Originally Posted by 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?

3. Good post? |

## Re: Combinations

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

4. Good post? |

## Re: Combinations

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

5. Good post? |

## Re: Combinations

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

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

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