AnkitK wrote:
Thankzz to all.
But i am still confused about the logic behind below concept . Dear Fluke pls elaborate :
"x will be maximum when y is maximum. What is the maximum value for y
Max value for y = minimum of (40,70).
Because the number of students with brown hair is 40, the maximum number of students to have both brown hair and blue eyes can only be 40"
How can the number of students having neither brown hair nor blue eyes be maximized?
If the overlap between blue eyed guys and brown hair guys is increased; means the number of students having BOTH blue eyes and brown hair be increased. In terms of the set \(n(A\cap B) is increased\)
For simplicity;
Ex: Say there are 4 students(a,b,c,d);
1 student has brown hair and 2 students have blue eyes;
Stem says: 1 student has brown hair and 2 students have blue eyes;
minimum number of students who neither have blue eyes nor brown hair
Say
a-> brown hair (1 student has brown hair)
b,c-> blue eyes (2 students have blue eyes)
Number of students to have both blue eyes AND brown hair = 0
d-> doesn't have either blue eyes or brown hair. Count: 1
Maximum number of students who neither have blue eyes nor brown hair
Say
a-> brown hair & blue eyes
b-> blue eyes
Number of students to have both blue eyes AND brown hair = 1(a)
c,d-> doesn't have either blue eyes or brown hair: Count: 2
Mathematically; you can see;
To find the maximum number of students that have neither. We need to have maximum possible number of students that can have both.
Why the maximum number of students that have both can't be greater than the minimum of the two.
Maximum number of students having both = min(Students with brown hair, Students with blue eyes)
Students with brown hair = 1
Students with blue eyes = 2
Can the number of students having both be "2". NO... if out of 4 students; only 1, just ONE student has brown hair, how can there be 2 students with brown hair and blue eyes. It is clearly mentioned in the question;
Likewise;
Because the number of students with brown hair is 40, the maximum number of students to have both brown hair and blue eyes can only be 40". If only 40 students have brown hair, how can 41 students have both. It can never be any number more than 40.
Thus the maximum number of intersection is always the minimum of two counts.
Now, if 40 students have both blue eyes and brown hair and question says there are 70 students to have blue eyes. There are another 30 students who have blue eyes but no brown hair.
Total becomes 70: 40(Both brown and blue), 30(just blue).
Total students=100
Neither = 100-70=30
Coming to the find the minimum number of intersection;
if there are 40 students - brown hair
70 - blue eyes
it becomes 40+70=110. But, the total strength is 100. Thus, 10 students have both blue and brown, but 0 zero students who have neither.
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75% (01:14) correct 
