THE LAST FIFTH GRADE OF EMERSON ELEMENTARY

THE LAST FIFTH GRADE OF EMERSON ELEMENTARY
April 12, 2016

Sunday, September 6, 2015

Probability ... none.

Hey, Writerly Friends and Math Brains.

I was working on a new MG novel this morning and came upon (are you sitting down?) a MATH problem.

It's not that I'm bad at math. It's just that my interest stopped at Algebra, because Algebra is all about logic. And this problem is beyond my ability because it involves (really, sit down now) probability. Probability was nearly the death of me in high school. Who knew I would need it as a writer.

Here is the problem. I'm hoping someone out there likes a good probability challenge and can figure it out.

There are twenty-four children in sixth grade at a boarding school, twelve boys and twelve girls. At mealtimes, they sit in groups of four at six tables. The seats are assigned and rotating. The school's administrators want each student to sit in as many different groupings as possible before they eat a meal with the same foursome.


Image: Capital OTC

Question: How many meals would it take for a student to sit in the exact same group of four?

Challenge question: If the tables always had two boys and two girls, how would that affect the answer?


UPDATE!

Thanks for your feedback, guesses, and reminders about what "!" means in math, everyone. (It's a factorial. Remember those?)

Based on responses in the comments and via Twitter, here's what you came up with. I'm a writer ... of course, I have follow up questions.

Question/Answer: It turns out, the tables don't really matter. What's important is 24 students grouped in fours. The equation is 24!/(24-4)! x 4! and 10,626 is the answer. Thanks to Jen Maschari and her husband Kurt, who researched the question and chimed in with an answer first. I'm really impressed with how many people had this right.

My follow up question is this: Is this number the number of possible combinations for *all* the students? If so, how often would one particular grouping of four meet during those 10,626 times?

Challenge question/Answer:
According to the amazing author/mathlete Marieke Nijkamp, the equation for this one is: (12!/(12-2)! x 2!) x (12!/(12-2)! x 2!). I have varying answers from the crowd on this one. Anyone care to work it through?

I have a follow up question that may turn parts of the question into a red herring. If you are a girl and one of the other girls in the class is your best friend, how often can you expect to eat a meal with her?

Would it help to assign the girl and her friend names? Girl X and Girl Y -- keeping it mathy.

Thanks also to my commenters: Tabatha Yeatts and her son Dash, June Smalls, Linda Baie, Jone MacCulloch, Sue Poduska. I appreciate your math brains. And on Twitter, thanks for helping out Vicki Coe, Mike Grosso, Dee Romito, and Abby Cooper. Am I the only author who's not a math whiz?

You can buy it here.

7 comments:

Tabatha said...

Dash is not sure, but he thinks it is 1,771 days for the original question. (There are 10,626 possible foursomes, divided by six is 1,771.) For the challenge question, he thinks it's only 726 days. (There are 4,356 possible foursomes.)

June Smalls said...

This was taken off a calculator where the answer was given. I believe that you are looking at 10626 different 'groups' before you can sit with all the students in all possible groups. (I got that number off a lotto website. I didn't do the math myself.

Directions: Apply the combination formula to solve the problems below.
Problem 1) In a class of 10 students, how many ways can a club of 4 students be arranged?
Answer
Combination Formula Applied
210

http://www.mathwarehouse.com/probability/combination.php

This was a fun way to get my brain working in the morning but without a scientific calculator I'm pretty useless.

Linda Baie said...

I got a slightly different answer, including not counting pairs in a math problem that are the same - AB or BA for example. There are 66 pair possibilities with just the boys and/or the girls. Combining them (as twosomes) would create 2,145 days of different foursomes. I'll keep checking, but have to go out.

Jone MacCulloch said...

If this were my WIP, I would b .doing what you're doing: asking for help. This is not my strength. It's sounds like you have answers.

Sue Poduska said...

10,626 and 4,356 are correct. Not sure why Tabatha is dividing by 6. There are still as many possible combinations no matter how many tables there are. It's a basic Bayesian combination. (24 choose 6, (12 choose 2) squared)

Author Amok said...

Thanks, everyone. I'm amazed by how many people were up for this challenge.

"Basic" problem: 10,626 seems to be the winner. My follow up question is this: Is this number the number of possible combinations for *all* the students? If so, how often would one particular grouping of four meet during those 10,626 times?

Challenge problem: I have a follow up question that may turn some of the information into a red herring. If you are a girl and one of the other girls in the class was your best friend, how often could you expect to eat a meal with her?

Would it help to assign the girl and her friend names? Girl X and Girl Y -- keeping it mathy.

Tabatha said...

Glad you got the correct answers! It's my fault that Dash divided it by 6, I expect, because I thought the number was awfully big. I thought you wanted it for something to happen in your book and it would never get around to happening if it took that many days!