Posted by **Mike** on Tuesday, February 19, 2008 at 6:52pm.

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?

- Math -
**Guido**, Tuesday, February 19, 2008 at 7:11pm
87 x 86 x 85 x 84 x 83 = ???

- Math -
**Reiny**, Tuesday, February 19, 2008 at 7:23pm
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

- Math -
**Mostly**, Sunday, February 10, 2013 at 7:20pm
4,433,982,840

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