SOLUTION: out of 136 students 67 read tribune, 56 read punch,40read the nation. 5 read all the three,12 read punch nation,9read tribune and nation.how many students read tribune and punch.

We have the set A of 136 students and three its subsets T (67 elements), P (56 elements), and N (40 elements).

Let us denote TP the intersection of T and P for brevity; denote TN the intersection of T and N; and denote
PN the intersection of P and N.

Also let us denote TPN the intersection of all the three subsets T, P and N.

We know that the subset T consists of 67 elements, the subset P consists of 56 elements and the subset N consists of 40 elements.

We are also given that PN consists of 12 elements, TN consists of 9 elements, and TPN consists of 5 elements.

The question is: how many elements are in TP?

For any given subset X, let us use the symbol |X| for the number of elements in X.

Then there is a remarkable equality, which connects the number of elements in the set A; in its subsets T, P and N;
in their intersections TP, TN and PN; and in the subset TPN:

|A| = |T| + |P| + |N| - |TP| - |TN| - |PN| + |TPN|.     (1)

Now look how this equality will help us to solve our problem.
Simply substitute the known values into this equality. You will get

136 = 67 + 56 + 40 - |TP| - 9 - 12 + 5.                 (2)

|TP| is the only unknown in this equation, and you can easily find it by isolating.

|TP| = 11.


Actually, the equality (1) is valid for any finite set A and for any three its subsets T, P and N that jointly cover the set A. 
The proof is in this dialog:

Physicist: Having the goal to calculate the number of elements in A, let us sum up the amounts of elements in subsets T, P and N:

   |A| = |T| + |P| + |N|.                               (3)

Mathematician: Hmmm, interesting . . . But it is not exactly right, because the subsets T, P and N have intersections.
               By adding up the amounts of elements according (3), we count twice the elements of each intersection TP, TN and PN.

               So, it works only as the "first approximation" of the number |A|.

Physicist: Agree. But we can easily fix the situation. Simply, take off the amounts of elements in the intersections, like this

   |A| = |T| + |P| + |N| - |TP| - |TN| - |PN|.           (4)

Mathematician: Hmmm, a good idea . . . But it is still not exactly right, unfortunately. 
               Notice that when we sum up (3), we count the elements in the tripled intersection TPN three times.
               In the next step (4), we take off these elements of the intersection TPN three times. 
               It means that the elements of the intersection TPN do not contribute at all to the value of |A| in formula (4).

Physicist: Yes, you are right.
               But we can make one step forward and add the amount |TPN| to the right side of (4).
               Then we will have 

   |A| = |T| + |P| + |N| - |TP| - |TN| - |PN| + |TPN|,

               exactly as the formula (1) says.

After completing the proof (it is really completing and it is really the proof) Mathematician and Physicist went for lunch.

(Alternative version: they wrote this text and submitted it to the forum www.algebra.com)