Katie has 9 employees that she must assign to 3 different projects. If

Your point does make sense Karun.

C(9,3) makes the selection of first group of 3 from a set of 9, and then C(6,3) makes a selection of the second group of 3 from the set of remaining 6 and then C(3,3) makes a selection of the final group of 3 from the remaining 3 (trivial case obviously).

However these three groups can be allocated to the three projects in 3! ways.

I think we should have a 3! multiplier to the answer.

What say guys? Any ideas on this one?

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.

Alternative approach:

9!/(3!*3!*3!) = 1680

We know there are a total of 9! possibilities without any groupings/restrictions.

Applying groupings/restrictions, there are 3 groups that must be made of 3 different employees (3!*3!*3!)

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.

EE Combos for (Project 1 * Project 2 * Project 3) / project permutations (corrective measure to eliminate over-counting)
3C9*3C6*3C3/3P2
=1680

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

\(9C_3*6C_3*3C_3*3 projectsC_3groups\)

Bunuel or some Quant guru, is the logic of the final term correct (Assigning 3 projects to 3 groups).

What you have done is CORRECT because 3C3 is 1 anyways

Why do we not divide by 3! when in this problem: https://gmatclub.com/forum/in-how-many- ... 01722.html

we divide by 3! ?

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

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

VeritasKarishma @expertglobal What is the problem in this approach. By doing 9c3,6c3, and 3c3 we are making a group of 3 person each .
But to take care of 3 different projects, multiplication with 3c1, 2c1, and 1c1 was done.

Please help !

Take a simpler case:
Case 1: You have 3 people and 3 distinct seats. How do you allocate? For seat1, you select in 3C1 ways. For seat2, in 2C1 ways and for seat 3, in 1C1 ways.
So 3C1 * 2C1 * 1C1

Take another case:
Case 2: You have 3 people and 3 distinct projects to allocate them. How will you do it? This is exactly like Case 1. Whether you have 3 distinct seats or 3 distinct projects or 3 distinct rooms or 3 distinct positions, doesn't matter. For first seat/project/room/position, you select in 3C1 ways. For second in 2C1 ways and for third in 1C1 ways. Basically the selection is made for some distinct spot in 3C1 ways, whatever that distinct spot may be.


You need to select 3 groups of 3 people each for 3 distinct projects.

Here again, you have 3 distinct projects so 3 distinct spots. We select 3 people for project 1 in 9C3 ways. The project 1 now allocated.
For project 2 in 6C3 ways. For project 3 in 3C3 ways.
No more arrangements are required.

Nirmesh83 karun_aggarwal

It is a case of selecting 3 persons in each group for 3 groups. Let's imagine Group A, Group B, and Group C

1. At the beginning, for selection of 3 persons in Group A, we have 9 persons left. Thus, number of ways in which 3 persons in Group A will be selected are 9C3.

2. After choosing 3 out of 9 persons, we are left with 6 persons. Hence, for selection of 3 persons in Group B, we have 6 persons left. Thus, number of ways in which 3 persons in Group B will be selected are 6C3.

3. Finally, After choosing 6 out of 9 persons for Group A, and Group B, we are left with 3 persons. Hence, for selection of 3 persons in Group C, we have 3 persons left. Thus, number of ways in which 3 persons in Group C will be selected are 6C3.

The number of ways in which these 9 people will be placed on 9 seats in all three groups is = Number of Ways in Group A (x) Number of Ways in Group B (x) Number of Ways in Group C

Thus, answer choice B is correct.

Bunuel KarishmaB
Why we aren’t we dividing by 3! To me it seems above linked question and this one are same... plz explain someone....I am totally puzzled.

Posted from my mobile device

The two questions are not the same.

Imagine a group of 9 people standing together (A, B, C, D, E, F, G, H, I).
You split them into 3 groups of 3 people each (say A,D,I and C,G, H and B, E, F).
This is your original question.

Now imagine you split the 9 people into 3 groups of 3 and assign these groups to 3 different spots on the field - spot 1, spot 2 and spot 3.
Now if A,D,I go to spot 1, it is different from if A,D,I go to spot 2.
The 3 groups need to be arranged in 3 spots which can be done in 3! ways.
That is why in the linked question, you need to multiply by 3!

This is the application of the basic counting principle discussed here: https://youtu.be/LFnLKx06EMU

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