Question

A) in a group of 2,000 people, must at least 5 have the same birthday? why?
b.show that if seven integers are selected from the first 10 positive integers there must be at least two pairs of these integers with the sum 11.

Answer 1

a) The answer is yes. 
Given 365 people, two at least must have the same birthday (obvious).
Now multiply the above number by 5 we get : 1825.
Since $1825\leq2000$ then 5 at least will have the same birthday. 
b)

Think of the number 11 and how many manner we can sum to 11:
1+10=11
2+9=11
3+8=11
4+7=11
5+6=11
So there is five ways. Suppose we draw one number. There is no way to draw 7 numbers without obtaining a couple forming 11.