Solving Permutation Problem: 7 Mech, 6 Civil & 5 Elec for 9 Presentations

[etotheix]

Homework Statement

If the class contains 7 mechanical, 6 civil and 5 electrical and we require that the 9 individuals who give their presentations on day 1 must include 3 mechanical, 3 civil and 3 electrical, how many different orders of presentation are there for day 1?

Homework Equations

The Attempt at a Solution

It looks like we have to combine 3 permutations together, taking only 3 students from each group.

I have no clue how to do this, I have tried reducing the problem to :
2 mechanical, 2 civil and 2 electrical and 1 per group must give a presentation.

I computed by hand the different orders of presentation, but then I don't know what formula to use to arrive at the same answer.

So If we have :
Mechanical students : A, B
Civil : C, D
Electrical : E, F

Then the different presentations on day 1 would be :

A, C, E
A, C, F
A, D, E
A, D, F
B, C, E
B, C, F
B, D, E
B, D, F

8*2*3 = 48 possible presentation order in this simple case, am I right?

 

[lanedance]

i would read and think about it a little different as follows

well first deal with the order of presentations in terms of discipline only, not considering who gives them... you have MMMCCCEEE, any arrangement of this will give a valid presentation schedule, so how many different ways to arrange this sequence are there?

so say you have a set sequence eg/ MECECCEMM, how can you arrange the different students...?

consider first the Ms, how many different ways can you arrange 9 distinguishable mech students through the given 3 Ms?

 

[lanedance]

I have no clue how to do this, I have tried reducing the problem to :
2 mechanical, 2 civil and 2 electrical and 1 per group must give a presentation.

I computed by hand the different orders of presentation, but then I don't know what formula to use to arrive at the same answer.

So If we have :
Mechanical students : A, B
Civil : C, D
Electrical : E, F

Then the different presentations on day 1 would be :

A, C, E
A, C, F
A, D, E
A, D, F
B, C, E
B, C, F
B, D, E
B, D, F

8*2*3 = 48 possible presentation order in this simple case, am I right?

so this simple case reduces to:
choose 1 student from 2 = 2!
do it for each of the 3 disclipines = (2!)^3
now count the ways to arrange the 3 discliplines = (2!)^3. (3!) = = (2)^3. (3.2) = 8.(3.2)

which agree with your value