SOLUTION: You can answer any 18 questions from the 23 questions on an exam. In how many different ways can you choose the 18 questions, assuming that the order in which you choose the questi

Algebra ->  Permutations -> SOLUTION: You can answer any 18 questions from the 23 questions on an exam. In how many different ways can you choose the 18 questions, assuming that the order in which you choose the questi      Log On


   

Question 665627: You can answer any 18 questions from the 23 questions on an exam. In how many different ways can you choose the 18 questions, assuming that the order in which you choose the questions is irrelevant?
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
The answer is the number of Combiations you can have by choosing a group of 18 from a total population of 23.
(1) 23C18 is the symbol we use, it is evaluated as
(2) 23C18 = 23!/((23-18)!*18!) or
(3) 23C18 = 23!/(5!*18!)
Using your calculator (find the ! key for factorial) to get
(4) 23C18 = 33,649,. If you don't have ! key (3) simplifies to
(5) 23C18 = (23*22*21*20*19)/120
You also want to know about a caterer picking meal.
For the appitizers we have 3C2 = 3
For the main dish we have 11C7 = 11!/94!7!) = 330
For dessert we have 4C2 = 6
The total number of choices is 3*330*6 = 5940