source : MGMAT CAT

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
13
10
9
8
7

I know there is explanation provided for this but I am not getting the steps.

I started from inside out. There are 3 thhat are in all three classes so

a = number of students in both History and Math
b = number of students in both History and English
c = number of students in both Math and English

For brevity I will only show the first step because herein lies my confusion.

Students in Math Only = 25- (a -3+3+b-3)

The explantion says Students in Math = 25 -(a+b+3) and this is the part that I am missing. Why would we not subtract 3 from the number of students that have both classes?

Can someone please shed some light? Thanks

This question is already dealt with on the forum.

one extremely efficient way to attack this problem is to use the following formula, which works for all combinations of three sets a, b, c:

# of items total = (a + b + c) - (ab + ac + bc) + (abc)

in the formula, 'a' stands for the # of items in set a (regardless of whether they are in other sets as well), 'ab' for the # of items that are in both sets a and b (again, regardless of whether they're in c), and so forth.

note that you cannot use this formula if you are given information with exclusivity (for instance, the number of items in set a and b but not in set c).

in this problem, you are looking for the value of (ab + ac + bc - 3abc). notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc'). since you already have abc = 3, all you need is the value of (ab + ac + bc).

using the formula,
68 = (25 + 25 + 34) - (ab + ac + bc) + 3
(ab + ac + bc) = 19
answer = 19 - 3(3) = 10

you can also solve this problem with venn diagrams and associated reasoning. hit up this thread again if you want more details about that.

How do you get 3(3) in the final step of the problem?
''answer = 19 - 3(3) = 10''

quoting from the relevant part of that solution:
notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc')

i.e., you don't want those items at all, but they've "accidentally" been counted three times in the existing part of the calculation. therefore, if you subtract them out three times, you're good.