Of the students in a certain class, 10 had taken a course in

A union B union C = A + B + C - AB - BC - AC + ABC
So
20 = 10 + 11 + 14 - AB - BC - AC + 3
=> AB + BC + AC = 18

This is too basic 3 elements set// simply draw a diagram with three overlaping circles// it will be much clearer//

ab + bc + ca = 10+11+14-20-3
ab + bc + ca = 12

sorry, but this is not the correct answer ....

Can somebody help me understand what does ".... and 20 students had taken a course in only one of A, B, and C ... " as stated in the question mean?

ouch.................

total = x
a = 10
b = 11
c = 14
abc = 3
only a + only b + only c = 20
ab + ac + bc = ?

a+b+c= (only a + only b + only c) + (ab + bc + ca) - 2(abc)
10+11+14 = 20 + (ab + bc + ca) - 2(3)
ab + bc + ca = 35 - 20 + 6
ab + bc + ca = 21

Is the OA 18?

A+B+C = A (10) + B (11) + C (14) - AnB - AnC - BnC + AuBuC (3)
Since A+B+C (without overlap) = 20...

20 = 10 + 11 + 14 + 3 - (all the students taking two classes at once, let's call this x)
x = 10 + 11 + 14 + 3 - 20
x = 18

Is this correct???

AnB + AnC + BnC = A + B + C - 3AnBnC - A(only) - B(only) - C(only) = 10 + 11 + 14 - 9 - 20 = 3

Answer = 12

Please see diagram below

Total enrollment: 35.

A, B, C only 20.

2 or 3 courses make up 15 enrollments.

3 people enrolled 3 times. We want to get rid of them. -9.

6 enrollments. 3 people double-counted.

Different ans? above OA is wrong or rite?

10+11+14-2(3)-x=20

i m geting 9?

this seems good .. I think this question bit different than other questions that we usually see in tests..

Not really.

Of the students in a certain class, 10 had taken a course in A, 11 had taken a course in B, and 14 had taken a course in C. If 3 students had taken a course in all of the A, B, and C, and 20 students had taken a course in only one of A, B, and C, how many students had taken a course in exactly two of A, B, and C?

Given that:
A = 10;
B = 11;
C = 14.

a + b + c = 20 (20 students had taken a course in only one of A, B, and C);
g=3.

Question:
How many students had taken a course in exactly two of A, B, and C: d + e + f = ?

Now, recall that
\(Total = A + B + C - (sum \ of \ EXACTLY \ 2-group \ overlaps) - 2*(all \ three)=\)
\(=10+11+14-(e + d + f)-2*3=29-(d + e + f)\) (check here: advanced-overlapping-sets-problems-144260.html - Theory, examples, detailed solutions).

We can get the total in other way too: \(Total = (a + b + c) +(d + e + f)+g=20+(d + e + f )+3=23+(d + e + f )\).

Equate these two: \(29-(d + e + f )=23+(d + e + f )\) --> \((d + e + f )=3\).

Answer: A.


Hope this helps.

[/quote]Not really.

Of the students in a certain class, 10 had taken a course in A, 11 had taken a course in B, and 14 had taken a course in C. If 3 students had taken a course in all of the A, B, and C, and 20 students had taken a course in only one of A, B, and C, how many students had taken a course in exactly two of A, B, and C?

Given that:
A = 10;
B = 11;
C = 14.

a + b + c = 20 (20 students had taken a course in only one of A, B, and C);
g=3.

Question:
How many students had taken a course in exactly two of A, B, and C: d + e + f = ?

Now, recall that
\(Total = A + B + C - (sum \ of \ EXACTLY \ 2-group \ overlaps) - 2*(all \ three)=\)
\(=10+11+14-(e + d + f)-2*3=29-(d + e + f)\) (check here: advanced-overlapping-sets-problems-144260.html - Theory, examples, detailed solutions).

We can get the total in other way too: \(Total = (a + b + c) +(d + e + f)+g=20+(d + e + f )+3=23+(d + e + f )\).

Equate these two: \(29-(d + e + f )=23+(d + e + f )\) --> \((d + e + f )=3\).

Answer: A.


Hope this helps.[/quote]

isnt it too long?

I mean cant i have formula..
I was doing with the formula..

A+b+c-3(all)-(sum of exactly two groups)
20= 14+11+10-2(3)-sum of exatly two groups..

I m geting 9..

where m wrong?

I think what you are doing wrong is not reading the question and solution carefully.

The total number of students is not 20: 20 students had taken a course in only one of A, B, and C, means that 20 = a + b + c.

Hope it's clear.

Exactly 1 + Exactly 2 + Exactly 3 = ALL = A + B + C - Exactly 2 - 2(in all A B and C)

20 + E2 + 3 = 10 + 11 + 14 - E2 - 2(3)
SOLVE:

E2 = 3

Bunuel: Why are you multiplying 2*(All three)?

This is in detail explained here: ADVANCED OVERLAPPING SETS PROBLEMS.

Hope it helps.

Got it right second time:

So 3 is overlapped 3 times,
The answer, x, is overlapped 2 times,
Area of 20 is not overlapped,
Sum is 35
=> 2x = 6, x=3.