Venn Diagram tyoe question

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 completly lost me there. I haven't learned about U or ^ yet.

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)

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.

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

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