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>
Use Pythagorean’s Theorem. a^2 + b^2=c^2. a^2 + 35^2 = 37^2 a^2 + 1225 = 1369 then subtract 1225 from both sides. a^2 = 144 then find the square root of both sides a = 12
you multiply it as you would regular numbers and then you count the places that the decimal is in front of. in this case it would be 2 numbers, then in the number you got by multiplying, move the decimal 2 places up
