# Venn Diagram tyoe question

1. Nov 3, 2012

### Psyguy22

These kind of questions always get to me and I don't know how to solve them.

Lets say that there are X many people that are in sports. Y of them are in soccer, Z of them are in cross country, and A of them are in basketball. And Y+Z+A>X

How would i find out how many people do two sports or all three?

2. Nov 3, 2012

### arildno

Let S(2,3) be those who practice two or three sports, (Y,Z),(Y,A),(Z,A) those practicing two sports, and (Y,Z,A) those practicing 3.
Then, S(2,3) equals the sum of those 4 disjoint groups.
Agreed?
Furthermore, let Y(0) be those ONLY playing soccer, and similarly for the 2 others.

Then, we have the equation:
S(2,3)+Y(0)+Z(0)+A(0)=X (*!*)

Now, we have, of course, Y(0)=Y-(Y,Z)-(Y,Z,A) and so on.

Now, inserting these into (*!*), we may simplify this to:

Y+Z+A-S(2,3)-(Y,Z,A)=X (!!!)

Therefore, in order to solve (!!!) for S(2,3) uniquely (knowing Y+Z+A and X), you need to know how many play 3 sports.

Obviously, (Y,Z,A) must be less than or equal to S(2,3)

If you know Y+Z+A-X=1, it follows immediately that (Y,Z,A)=0

Last edited: Nov 3, 2012
3. Nov 3, 2012

### Psyguy22

First, what do you mean by *!* is that some kind of factorial? And same thing with !!!? Other than those, I followed that pretty well.

4. Nov 3, 2012

### arildno

Those were NAMES I gave to my favourite equations. If you prefer to call them "Peter" and "Polly", by all means do so.

5. Nov 3, 2012

### paras02

Firstly find n(a^y)
then n(y^z) , n(a^z) and n(a^y^z)
The answer will be = n(a^Y)+n(y^z) + n(a^z) - 2*n(a^y^z)
where n(a^y) denotes no. of players who play both soccer and basketball,
n(a^y^z) denotes no. of players who play all the three games

Last edited: Nov 3, 2012
6. Nov 3, 2012

### Psyguy22

I'm sorry, you completly lost me there. I haven't learned about U or ^ yet.

7. Nov 4, 2012

### arildno

He is using ^ as the logical operator AND.
There are many ways to split up a Venn problem, hopefully, the approach I gave you made sense (even though I gave my equations names, but didn't inform you on that)

8. Nov 4, 2012

### Psyguy22

Ok. So now I'm trying to understand this more.
I just made up these numbers
There are 24 people.12 play soccer, 9 run cross, and 10 are in basketball. How many play two sports? How many play three?
I tried putting in Y(0)=12-(y,z)-(y,z,a) but Im not.sure how to simplfy that.

9. Nov 4, 2012

### arildno

Well, X=24, Y+Z+A=31
Thus, you have, by inserting in (!!!), and rearranging:
S(2,3)+(Y,Z,A)=7 (agreed?)

Now, this can refer to the following situations:
a) There are 7 players who play two sports, none playing all
b) There are 5 players who play two sports, and 1 playing all
c) There are 3 players who play two sports, and 2 playing all
d) There is 1 player who plays two sports, and 3 playing all

10. Nov 4, 2012

### arildno

In total, you have 70 unique arrangements satisfying the conditions you gave, with
36 unique arrangements of the a)-solution
21 unique arrangements of the b)-solution
10 unique arrangements of the c)-solution
3 unique versions of the d)-solution.

11. Nov 4, 2012

### Psyguy22

So this question has multiple answers?
Thank you for your guys help!

12. Nov 4, 2012

### arildno

With no further information given, yes.
In exercises, there will usually be additional information to specify down to unique solution.
You're welcome.