In how many ways can 19 students be divided into five groups, two groups of 5 and three groups of 3, so that each group studies a different topic?

Guest Sep 22, 2017

#1**0 **

In how many ways can 19 students be divided into five groups, two groups of 5 and three groups of 3, so that each group studies a different topic?

I will state right off that i do not know how to do this.

This would be my guess but it is probably wrong.

Choose 10 from the 19 This can be done in 19C10 ways

Then chose 5 from those 10. This can be done in 10C5 ways

But if you have one group of 5 you automatically have the other group of 5 so I halved this.

So far that is

19C10 * 10C5 / 2!

Then choose 3 from what is left 9C3 ways

Then choose 3 from what is left 6C3 ways

but it doesn't matter what order these are chosen in so that is divided by 3! ways

so that is

9C3 * 6C3 / 3!

Put that together and I get

(19C10 * 10C5) / 2! * (9C3 * 6C3) / 3!

(nCr(19,10)*nCr(10,5)/2)*(nCr(9,3)*nCr(6,3)/3! = 325 9095 840 ways

It is probably wrong.

If you find out how do do it ,for certain, could you please let me know :)

Melody Sep 23, 2017

#2**+1 **

I think that the answer is 19C5*14C5*9C3*6C3, whatever that works out to be.

This business of dividing by 2 and 3! when you have one group of five then the other group of five is determined, or for the groups of three it doesn't matter in what order they are chosen, is wrong.

There is a difference between splitting the 19 into two groups of five and three groups of three without distinction, and splitting them into groups where each group studies a different topic.

Consider, for example, a group of four students, call them A,B,C and D, split into two twos.

4C2 = 6, so we have 6 pairs, AB, AC, AD, BC, BD, CD.

However, when the first pairing is chosen, in effect the second pairing is also chosen, so we have just three (= 6/2) possible groupings, {AB, CD}, {AC, BD} and {AD, BC}. That's the situation if it doesn't matter which pairing comes first, that is, for example, {AB, CD} is the same as {CD, AB}.

Now suppose that the students are going into two different language classes, say the first pairing study French and the second pairing study German. Now, the order of the pairings makes a difference, {AB, CD} is different from {CD, AB}.

In the first case, A and B study French and in the second case they study German, (and the other way round for C and D).

If this is the situation, the original calculation of 4C2 = 6 stands.

(Notice btw that, for the 19 student case, as far as the calculation of the final number is concerned, it doesn't matter which group is chosen first, that is, for example, 19C5*14C5*9C3*6C3 = 19C3*16C3*13C5*8C5 = ... .)

BobP

Bertie Sep 25, 2017

#3**0 **

Thanks Bob I really appreciate your answer.

I shall give what you say a lot of thought.

Melody
Sep 25, 2017

#4**0 **

Of course, thanks Bob, my logic was fine I just didn't read the question properly.

I didn't realise that the 2 groups of five were studying different subjects.

**Thank you very much :))**

Even after you answered it took me ages to realise that I had just not read the question properly.

Could you please say if my answer would have been right if the there had been no different subjects involved, so that order would not matter. ?? I think it seems right but I am not at all sure ://

Melody
Sep 30, 2017