Show that in any group of 101 people there are at least two persons having the same number friends (it is assumed that if a person x is a friend of y then y is also a friend of x and we are only considering the friends amongst the group of people). how would you modify your proof if the number of people were not 101?

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.

first option

a function is a relation in which an input is paired with exactly one output.

so, let's find the right answer.

in our first option, our inputs are january, february, and march. the outputs are 30, 28, and 25. each input has it's own output. this is correct!

our second option we know is incorrect, because ms. valverade has two outputs, dogs and cats. this is not a function.

our third option is incorrect as well. the yellow square has two ouputs, the square and the triangle.

lastly, in our fourth option, we see the number 7 two times with a different output. therefore, this is not a function.

remember: each input must be paired with only one output!

c

step-by-step explanation: