Of the 45 students in a certain class, 30 joined math club,

It's 7.

a. Math + Physics : 25
b. Math + Physics + Biology = 8
From a & b ==>
Only Physics = 0 & Only Physics + Math = 17 --> I

a. Math + Biology = 12
b. Math + Physics + Biology = 8 --> II

From a & b ==>
Only Math + Biology = 4 --> III

From I II & III : Only Math = 30 - ( 17 + 8 + 4 ) = 1

a. Physics + Biology : 0 ;
b. Math + Physics + Biology = 8 ;
c. Math + Biology = 4

a, b , c together ==> Only Biology = 20 - ( 12 + 8 ) = 8

Total : 45
No Clubs = k

k = 45 - ( Only Math + Only Physics + Only biology + ( Only Math & Physics) + ( Only Math & Biology ) + ( Only Physics & Biology) + ( All three)

==> k = 45 - ( 1 + 0 + 8 + 17 + 4 + 0 + 8 ) = 45 - 38 = 7

oops

Only biology is : 20 - ( 8 + 4 ) = 8

The correct answer is 7

Total = 45

Math = 30
Physics = 25
Biology = 20

P&M = 25
M&B=12
P&B=8
P&B&M =8

Solving it traditionally:
30+25+20 - 25- 12 - 8 + 8 + None= 45
None = 7

However, I notice that you don't actually need to consider Physics students since all physics students are in Math. I wish I can draw, but basically, the physics circle is in math circle. Therefore, if you use just math and biology and m&b, then you should get the answer...
30+20 - 12 + None = 45
None = 7

nice work. btw, what you use to draw these pictures?

how do you know that p&b=8 ?

Since all P is in M, and you know that P&M&B = 8
This means P&M&B = P&B = 8

As usual , gr8! .
And yes I did the same way and got 7 .

Just want to share strategy about cracking Venn diagram problem.

Basically, I don't know any Venn formula. What I do is I keep track of how many items has been counted. The total always consist of counting everything ONCE!

Forget about None for now...

For example, if you are give two circle Venn diagram, you have total of A and B and the overlap (A&B). You know that if you add A and B, the overlap has been count twice, therefore, for formula will be:
Total = A+B - A&B

If you are given three circle Venn diagram of A, B, C, it is a bit more tricky, but same trick.
If you are given A&B, B&C, C&A, you know that A,B,C already count A&B&C three times. A&B count A&B&C one time. B&C count A&B&C one time. C&A count A&B&C one time. Therefore, to make everything count ONCE, you can come up with the following:

A+B+C - A&B - B&C - C&A + A&B&C = Total
A+B+C => count A&B&C three times
A&B => count A&B&C one time
B&C => count A&B&C one time
C&A => count A&B&C one time
From this, you know that the "counts" are all three and they cancel out and therefore, you must ADD A&B&C at the end to make everything count ONCE!

Now, some complicated problems mention of "exactly two". This means that you are given A&B, but it doesn't count A&B&C. Using the same trick, if you are given exactly A&B without A&B&C, B&C with out A&B&C, C&A without A&B&C, your formula becomes:

A+B+C - A&B - B&C - C&A - 2*A&B&C = Total

Why? Because A, B, C count A&B&C three times. You know that A&B, B&C, C&A don't count A&B&C at all. Therefore, to make everything count ONCE, you need to subtract 2*A&B&C to the equation.

If you know how to do the "count", you don't even need to remember any formula. I find a lot of problems that the formula will just make it confusing, though mostly tough problem. Using this trick, you won't make a mistake in Venn diagram type problem.

Using None is a cake walk at this point.

Cheers.

Ugghhh Venn's one of biggest weaknesses.

How do u know B+P=0???

MS Powerpoint! Then cut and paste it as a picture =)