Agroup of 100 people is divided into 2 teams with 45 people in team a and 55 people in team b. a) if 20 of the people are mathematicians and 80 are non-mathematicians, in how many ways can the teams be formed if all the mathematicians are in team a? b) in how many ways can the teams be formed if none of the mathematicians are in team a?

Team A) 45 people
Team B) 55 people

A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A. 
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =363,413,731,121,503,794,368
or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20

As you can, we get the same answer for both. 

B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = 5,790,061,984,745,3606,481,440
or 5.79 x 10^22

Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer

It is the top one. the answer is -7. that is smallest