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)!) =<span>363,413,731,121,503,794,368 </span>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)!) = <span>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</span>
