How to Solve Venn Diagram Questions with Multiple Possible Answers?

[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?

 

[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)

This CAN help you in a specific case:
If you know Y+Z+A-X=1, it follows immediately that (Y,Z,A)=0

 

[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.

 

[arildno]

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

 

[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

 

[Psyguy22]

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

 

[arildno]

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

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)

 

[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 I am not.sure how to simplfy that.

 

[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

 

[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.

 

[Psyguy22]

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

 

[arildno]

So this question has multiple answers?

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

Thank you for your guys help!

You're welcome.