Answer:
Step-by-step explanation:
Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.
Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved
Part ii: EVen if instead of 101, say n people are there same proof follows as
if different number of friends then they would be 0,1,2...n-1
If one person has n-1 friends then there cannot be any one who does not have any friend.
Thus same proof follows.
You use the FOIL method for this problem. you should get X^2-25
1. David Hilbert was the perfect example of a celebrated, established and meticulous mathematician
2. David Hilbert achieved grand successes and great changes in his personal life as well.
3.
David Hilbert married his second cousin, KŠthe Jerosch on October 12, 1892 and the following year, Franz, their son, was born.
Read more at http://www.thefamouspeople.com/profiles/david-hilbert-471.php#5q42OtAom0OcDjpD.99
