math help!?

Question

a club has 100 members. How many different pairs of members could be chosen to be co-presidents? Explain please!

Answer 1

100*99/2=4950

for the 1st co president, there are 100 to choose from. for the 2nd, there are 100-1=99
divide by 2 because of reversibility (member a & member d is the same as member d & member a).

Answer 2

Well, how many people could be the first co-president? 100. If you've already chosen the first co-president, how many choices are there for the second? 99. So there are 100*99 different ways to choose first and second co-presidents, however, it doesn't matter what order we choose them in. For each possible pair there are two different orders they could have been chosen, so we divide by 2. Thus the number of possible pairs is 100*99/2.

Answer 3

one pair =>2 members

So, we want to choose 2 members among 100 members

Generally we obtains k pairs to choose among n members

=> n!/ k!(n - k)!

n =100 and k =2 => 100!/ (100 - 2)! = 100!/98!

= (100 x 99 x 98!)/98!

= 9900 pairs of members to
be co-presidents.

Answer 4

How many Co presidents can be chosen in total?

Answer 5

If this is a tricky math question it would get another answer, however, in the real world a club would have By-laws which would specify the officers' positions (Pres, VP, Sect, Treas, etc) required and the amount of persons who would hold those positions, and usually the length of term, etc.
As for a math problem, time to go back to your textbook and learn the equation. Unfortunately, I don't do homework.

Answer 6

it is a problem of combination without repetition with n = 100 and r = 2
total = n! / (n! (n - r)! )
so the total numbers of pairs is 9900

Answer 7

how many r there who wants to be co-presidents??