Problem Solving

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

PLZZZ, help me

vdemidoff wrote:John has 9 employees that he must assign to 3 different projects. If three employees are assigned to each project and no one is assigned to multiple projects, how many different combinations of project assignments are possibble?

PLZZZ, help me

I think this is a combination questions not a permutation. anyways this would be my approach.

Selecting small group from a large one..
n!
_______
k! (n-k)1

9!
________
3! (9-3)!

9 x 8 x 7
_______
3 x 2 x 1

3 x 4 x 7 = 84 .. what is the Official Answer?

Correct answer is 1680.
How it can be?
the first step we have to find the number of ways of choosing 3 out of 9. 9C3 = 84, but then I have to find the number of choices 3 guys for 3 different projects the Number= 3*2*1=6
84*6=504

BUT ANSWER IS 1680

Correct answer is 1680.
How it can be?
the first step we have to find the number of ways of choosing 3 out of 9. 9C3 = 84, but then I have to find the number of choices 3 guys for 3 different projects the Number= 3*2*1=6
84*6=504

BUT ANSWER IS 1680

vdemidoff wrote:Correct answer is 1680.
How it can be?
the first step we have to find the number of ways of choosing 3 out of 9. 9C3 = 84, but then I have to find the number of choices 3 guys for 3 different projects the Number= 3*2*1=6
84*6=504

BUT ANSWER IS 1680

You're right and I'm wrong.. Thanks for the explanations...

the general form for such example is may be:
x+y+z=9 , where x, y, z is the number of the employee the x, y, z project.
the problem is to find the number of combination
xc9*yc8*zc7, because we don't want that more than one employee to be assigned to one project, so we need to substract from the previous combination.
you can substitute x=9-y-z
then (9-y-z)c9*yc8*zc7.
for simple illustration
x+y+z=4
we have 2c4*1c3*1c2+1c4*3c3*1c2+1c4*1c3*2c2.
I hope that will help
thanks for more suggestions

Maybe you are right, but your way does not lead to the right answer (1680)
abd it looks very complicated, but anyway Thanks a lot!!

9 different people should be assigned to 3 different projects.

Say you have 9 spaces. _ _ _ _ _ _ _ _ _.
You now need to arrange 9 people in 3 groups (say G1, G2, G3)

G1 G1 G1 G2 G2 G2 G3 G3 G3

Now we have 3 each for G1, G2, G3

now it can be computed as 9! / (3!*3!*3!) = 1680 (We are dividing by 3! thrice for repeating groups of G1, G2 and G3)

Hope this helps

How about this:

(1) choose 3 persons from the first 9 : 9C3 = 84
(2) choose another 3 persons from the remaining 6: 6C3 = 20
(3) only one other way to choose 3 people for the last group is left, so you can just:

1 * 20 * 84 = 1680

was matching this problem with one I saw earlier in the forum. I agree with papgust, we can view this problem as a trinomial factor n!/x!y!z!.
thank you for the comment

salar_snake wrote:How about this:

(1) choose 3 persons from the first 9 : 9C3 = 84
(2) choose another 3 persons from the remaining 6: 6C3 = 20
(3) only one other way to choose 3 people for the last group is left, so you can just:

1 * 20 * 84 = 1680

Ok, maybe I did not get the meaning of this task.
I should choose 3 guys from 9, than I have to find in how many ways I can arrange 3 groups using these 3 guys. Am I right?

salar_snake wrote:How about this:

(1) choose 3 persons from the first 9 : 9C3 = 84
(2) choose another 3 persons from the remaining 6: 6C3 = 20
(3) only one other way to choose 3 people for the last group is left, so you can just:

1 * 20 * 84 = 1680

Ok, maybe I did not get the meaning of this task.
I should choose 3 guys from 9, than I have to find in how many ways I can arrange 3 groups using these 3 guys. Am I right?

Ok, 9! / 3!*3!*3! works and I know how to do the caluculation but I don't understand how you get from the word problem to that formula. Could someone break it down for me a bit more? I mean I see how you got that but presented with a similar problem I may not understand how to apply it.

I solved this sum on my own and I also used the 9C3*6C3*3C3 method, but what confuses me is that this gives the no. of ways in which people are chosen for each project, but the order in which the projects get to choose ppl is not decided. For eg. if we name projects as A, B and C. Don't we also need to decide whether they will get to choose in the order ABC, ACB, BCA? This is what I thought when I solved the sum and therefore I multiplied (9C3*6C3*3C3) by 3. Because I think in this way the choices made by each would differ, for eg. if A gets to choose 1st it gets 9C3, if 2nd then 6C3 and if 3rd then 3C3. Please help me understand this...therefore in my ans i got 1680 x 3.