SOLVED THE QUESTION.
Okay this problem can be solved.I crunched my brain hard and devised a way.This can be done using combinatorics and sets.
Yep you are right combinatorics in sets!
Now ,when there are 2 sets (say A and B).There is just one intersection.You know how to calculate this?
I'll show.
In intersection of elements are two places.
No of ways for choosing first element : 2 [you could either choose A or B]
Now of ways for choosing second element: 1 [after you have chosen ,say A,you are left with just B]
So no of ways = 2*1=2.But wait A intersection B is same as B intersection A,so divide by 2.
No of intersections =1
Similarly in a 3 element set Venn diagram.
No of intersections of 2 elements : (3*2)/2! =3
No of intersections of 3 elements : (3*2*1)3!=1
Now for a 4 element set:
No of intersections of 2 elements : (4*3)/2!=6
No of intersections of 3 elements : (4*3*2)/3!=4
No of intersections of 4 elements : (4*3*2*1)/4!=1
Now,
In a 2 element set
Let A denote only A
B denote only B
and a denote intersection of A and B.
Then total no elements = A+B+a
Similarly in a 3 element set,
Total no of elements : A+B+C+a+b+c+d [ where a,b,and c denote intersection of 2 elements and d denotes intersection of all 3 elements]
Now IN A 4 element set
Total = A+B+C+D+ [a+b+c+d+e+f]+[g+h+i+j]+[k] [ arranged order of intersection]=40
a= A intersection B
b= A and C
c= A and D
d= B and C
e= B and D
f = C and D
g= A and B and C
h= A and B and D
i = A and C and D
j = B and C and D
k= A and B and C and D
Question is what is k?
Now stay here only if you want your brains to be FRIED!
k = 40 - [a+b+c+d+e+f] - [g+h+i+j] - [A+B+C+D]
A+a+b+c+g+h+i+k =37 [English] [Include all those that have A]
B+a+d+e+g+h+j+k = 31 [Spanish]
C+b+d+f+g+i+j+k = 28 [Spanish]
D+b+e+f+h+i+j+k =27 [German]
Adding all of them!!!!!!!
A+B+C+D+2[a+b+c+d+e+f]+3[g+h+i+j]+4k = 123
Substituting:
A+B+C+D+[a+b+c+d+e+f]+[g+h+i+j]+k =40 in the above equation:
[a+b+c+d+e+f]+2[g+h+i+j+k]+3k = 83
WHAT THE HECK!
YOU DIDN'T PROVIDE COMPLETE INFO.
I NEED no students in 2 classes and 3 classes!
I assume no of 2 students = 34
No of 3 students =17
[I WON'T LET MY EFFORT GO WASTE]
34+34+3k=83
3k=15
k=5
or no of students who study all classes = 5
THIS IS NOT A GMAT QUESTION.
THIS IS ATLEAST A 5-10 MINUTE QUESTION
Featured Course
