This Week’s Math Problem: The Answer to the One About Being a Player
February 5, 2010
Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!
This Week’s Math Problem:
You are a player who intends to schedule one date every day of the week, plus an extra on Saturday. If you have already made plans for Sunday through Wednesday, and have an additional 10 people on a waiting list to go out with you, how many different social schedules could you arrange for the week?
The Answer: 5,040
If you’ve already made plans for Sunday through Wednesday, then you have four date-slots left to fill: Thursday, Friday, Saturday afternoon, and Saturday evening. Since you have 10 choices available and “order matters” (that is, going out with A on Thursday, B on Friday, and C and then D on Saturday is a different social schedule then going out with A, B, C, and D in a different order), we simply multiply 10 x 9 x 8 x 7 to get the answer, 5,040.
For those who haven’t seen a problem like this before, the logic is: you have 10 choices for the first date. But once you’ve chosen someone, you have only 9 options left for the second date, etc. And, of course, the Fundamental Counting Principle tells us that we multiply to get the final answer.
Note: A plethora of people got this in the comments! Accordingly, next week’s problem will be harder! See you Monday!