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

A. 252
B. 1,680
C. 2,340
D. 362,880
E. 592,704

i get yr points guys,,,but since there are three differet projcts also to choose from so i guess it should be

(9c3*3c1)*(6c3*2c1)*(3c3*1c1)

The answer should be \(9C_3*6C_3*3C_3\)

There is no need for a further factor of 3! since the order has already been taken into account. In fact, if the question had also stated that the three projects are indistinguishable, then it would have been required to divide the expression by 3! to account for having already taken the order into account.

The formula is nCr * (n-r)Cr * (n-2r)Cr * .... (n- (p-1)r) C r
where n is the total number of persons, r is the number of persons to be selected for each one of p groups

= 9C3 *6C3*3C3 = 1680

Explanation: Initially there are 9 persons. You can select any 3 persons to assign to project 1 or project 2 or project 3. Assume you assign them to project 1. Now there are 6 persons left and you can assign any 3 of them to project 2 or project 3. Assume you assign them to project 2. Finally there are 3 persons left with all those 3 persons to be assigned to project 3.

In the above you chose project 1 first, project 2 next and project 3 last. This project selections can be done in3! ways. Suppose had you assigned them to project 2 first then to project 1 and then to project 3 the same set of arrangements would happen. So you do not have to multiply by 3!. Generally speaking when you take care that any group of r persons may be assigned for each slot/task in arriving at the answer then this above ordering is not required.
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The ways to select three employees out of 9 for first project = 9C3

The ways to select three employees out of Remaining 6 for Second project = 6C3

Third Project already has the group of three to be assigned to it so need to select that

So total ways to assign 3 employees to each of 3 projects = 9C3 * 6C3 * 1 = 1,680

Answer: Option B
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What you have done is CORRECT because 3C3 is 1 anyways
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The first 3 employees can be selected in 9C3 = (9 x 8 x 7)/3! = 3 x 4 x 7 = 84 ways.

The next 3 employees can be selected in 6C3 = (6 x 5 x 4)/3! = 20 ways.

The final 3 employees can be selected in 3C3 = 1 way.

Therefore, the total number of ways the 3 projects can be assigned is:

84 x 20 x 1 = 1,680

Answer: B
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