1. ## a wee puzzle

A club receives 92 applications for 49 available spaces, the applications are received from 3 different groups, group 1 consists of 68 applicants, group 2 consists of 16 applicants, group 3 consists of 8 applicants.
All of the applicants are allocated a number 1-92, placed into a hat & drawn out randomly
What are the odds of 1/2 of each group being successful applicants?

2. ## Re: a wee puzzle

With 92 applicants, divided into groups of 68, 16, and 8, "1/2 of each group" would be 34+ 8+ 4= 46, not 49, so that is impossible. Did you mean "at least half"?

3. ## Re: a wee puzzle

Apologies, yes, at least 1/2

4. ## Re: a wee puzzle

Originally Posted by oldno7
A club receives 92 applications for 49 available spaces, the applications are received from 3 different groups, group 1 consists of 68 applicants, group 2 consists of 16 applicants, group 3 consists of 8 applicants.
All of the applicants are allocated a number 1-92, placed into a hat & drawn out randomly
What are the odds of 1/2 of each group being successful applicants?
As corrected to read "at least one-half of each group".
This calculation is for one half plus one of each group. SEE HERE.
I chose to do that one because it nicely assigns all available memberships.
But you can see there are many possible assignments for "at least" half.
Assigning exactly half lives three memberships unassigned. Those three can be assigned to three groups in ten ways.

5. ## Re: a wee puzzle

Since 49- 46= 3 this is just a matter of determining how many ways you can allocate 3 slots to 3 different groups. There are 3 ways to give all three slots to one group. There are 6 ways to allocate 2 of the slots to one group and one to the other. There are 3!= 6 ways to one slot to each group.

6. ## Re: a wee puzzle

Originally Posted by HallsofIvy
Since 49- 46= 3 this is just a matter of determining how many ways you can allocate 3 slots to 3 different groups. There are 3 ways to give all three slots to one group. There are 6 ways to allocate 2 of the slots to one group and one to the other. There are 3!= 6 ways to one slot to each group.
If we count the number of ways to arrange the string of two bars and three stars, $||***$, that gives the number of ways to place three memberships into three groups.
That is the arrangement $*|~|**$ would mean group one gets one, group two gets none, and group three gets two.
As I said above it is $\dfrac{5!}{3!\cdot 2!}=10$ ways.