Dividing people into groups (1 Viewer)

[SuxMATH]

How many options are there for 25 students to divide into groups of 3,4,5 and 6?

My solution was 25C6 x 19C5 x 14C4 x 10C3 / 4! but not sure if right or wrong. Someone please clarify.

 

[Drsoccerball]

How many options are there for 25 students to divide into groups of 3,4,5 and 6?

My solution was 25C6 x 19C5 x 14C4 x 10C3 / 4! but not sure if right or wrong. Someone please clarify.

Why you dividing by 4! ?

 

[SuxMATH]

Why you dividing by 4! ?

Since the ordering of groups does not matter, and since there are 4! ways of ordering the four groups? Not sure.

 

[Drsoccerball]

Since the ordering of groups does not matter, and since there are 4! ways of ordering the four groups? Not sure.

you dont divide by 4!

 

[Drsoccerball]

If the groups where unable to be differentiated then you would divide by 4 not 4! but these groups have different numbers so therefore you can see the difference

 

[Ekman]

If the groups where unable to be differentiated then you would divide by 4 not 4! but these groups have different numbers so therefore you can see the difference

Actually it is still 4!, because there will be 4! ways of there being an overlap.

 

[SuxMATH]

Actually it is still 4!, because there will be 4! ways of there being an overlap.

So no dividing?

 

[sharoooooo]

just do it for each group of 3,4,5,6 and then add it all together?

 

[Ekman]

So no dividing?

Yes, there is no need to divide.

 

[InteGrand]

What is this Q asking exactly? The wording isn't clear. Does it mean how many ways can we divide them into a groups of 3, b groups of 4, c groups of 5, and d groups of 6, where 3a + 4b + 5c + 6d = 25, a, b, c, d (non-negative) integers?

 

[braintic]

Of course, we are really dividing into groups of 3, 4, 5, 6 and 7.