Math
posted by Mike on .
In a raffle, 5 winners get to choose from 5 prizes, starting with the first name drawn. If 87 people entered the raffle, how many ways can the winners be arranged?

87 x 86 x 85 x 84 x 83 = ???

Your question, "If 87 people entered the raffle, how many ways can the winners be arranged?"
seems to imply that the prizes do not enter the picture.
The winners can be arranged in 87*86*85*84*83 or 4433982840
If you imply that each of them could have picked a different prize, let's argue this way:
The first winner could have been picked in 87 ways, and each of those could have picked one of the 5 available prizes, so
87*5 ways
the second winner could have been picked in 86 ways, assuming one cannot win again, and each could have picked one of the remaining 4 prizes, so 86*4 ways
so the total number of ways arranging the 5 winners each with a different prize
= 87*5*86*4*85*3*84*2*83*1
= big 
4,433,982,840