My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Thanks Tree1Thanks
Reply
 
LinkBack Thread Tools Display Modes
August 25th, 2016, 05:09 AM   #1
Member
 
Joined: Oct 2014
From: Mars

Posts: 30
Thanks: 0

Question Number of combinations?

Hi all,

I have a very confusing problem to solve. No idea how to do it

Imagine there is a list of 300 items.
You must pick 10 items from that list. Other people must do the same.

Problem 1: What are the chances that any 2 items out of the 10 that you picked will match the same quantity (2) of another person?

and

Problem 2: What are the chances that any 2 or more items out of the 10 that you picked will match the same quantity of another person?


Thanks in advance,
A.
Apple30 is offline  
 
August 25th, 2016, 06:46 AM   #2
Newbie
 
Joined: Dec 2015
From: Connecticut

Posts: 21
Thanks: 4

Math Focus: I love to hate all of them equally
I'll start by answering Part 1. If I can solve the second part, I'll include it in another post.

The chances of us constructing a list with exactly two matches is equal to the number of lists with two matches divided by the total number of possible selections. (It is helpful to think that our partner's choices are predetermined. After all, they have to pick something.) There are $\binom{300}{10}$ total ways to select 10 items from a list of 300. This is the denominator of our probability fraction.

The numerator is the number of lists with exactly two matches. It is helpful to envision us making our choices with the other person's choices in mind. Our two matches should "come from" our partner's list, and the rest should come from the complement of their list and the original.

There are $\binom{10}{2}$ ways to choose the matching elements from our partner's list. There are $\binom{290}{8}$ ways to choose the eight non-matching elements to complete our list.

The chances of us building a list with two matches are therefore $$\frac{\binom{10}{2} \binom{290}{8}}{\binom{300}{10}}$$. By my calculation, that's about 3.622%.
Brachydactyloid is offline  
August 25th, 2016, 07:17 AM   #3
Newbie
 
Joined: Dec 2015
From: Connecticut

Posts: 21
Thanks: 4

Math Focus: I love to hate all of them equally
The second part isn't nearly as intimidating as your bolded "or more" would suggest. It's simply a lengthy "or" statement (two or three or four ... or ten matches). Of course, we can't have more than ten matches. Each option is solved in the same manner as the previous part.

Say we want a list with $k$ matches. We use the exact same technique as above, yielding $$\binom{10}{k} \binom{290}{10 - k}$$ total ways to produce this list. We're looking for $k=2$ or $k=3$ or ... or $k=10$, which, in probability, is just a summation.

Our numerator is $\sum_{i=2}^{10}\binom{10}{i}\binom{290}{10-i}$, the sum of all desirable matching values. The denominator is the same as before.

The answer is $$\frac{\sum_{i=2}^{10}\binom{10}{i}\binom{290}{10-i}}{\binom{300}{10}}$$
Brachydactyloid is offline  
August 25th, 2016, 07:29 AM   #4
Newbie
 
Joined: Dec 2015
From: Connecticut

Posts: 21
Thanks: 4

Math Focus: I love to hate all of them equally
Quote:
Originally Posted by Brachydactyloid View Post
The answer is $$\frac{\sum_{i=2}^{10}\binom{10}{i}\binom{290}{10-i}}{\binom{300}{10}}$$
This is about 3.907%, which seems reasonable to me.
Brachydactyloid is offline  
August 25th, 2016, 11:23 AM   #5
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,112
Thanks: 1002

Quote:
Originally Posted by Apple30 View Post
Imagine there is a list of 300 items.
You must pick 10 items from that list. Other people must do the same.

Problem 1: What are the chances that any 2 items out of the 10 that you picked will match the same quantity (2) of another person?
What d'heck does that mean?!

I pick 10 numbers from 300 numbers, say 1 to 300;
you pick in exactly same fashion.
I picked: 7 34 63 89 134 167 186 207 255 267
If you pick 89 and 167 (along with any other 8 ),
then you've "matched" me?????????????????
Denis is online now  
August 25th, 2016, 02:20 PM   #6
Newbie
 
Joined: Dec 2015
From: Connecticut

Posts: 21
Thanks: 4

Math Focus: I love to hate all of them equally
Quote:
Originally Posted by Denis View Post
What d'heck does that mean?!

I pick 10 numbers from 300 numbers, say 1 to 300;
you pick in exactly same fashion.
I picked: 7 34 63 89 134 167 186 207 255 267
If you pick 89 and 167 (along with any other 8 ),
then you've "matched" me?????????????????
That's how I interpreted it. Although, based on his phrasing, his other selections couldn't just be any other eight; they would have to be options not included in your list.
Brachydactyloid is offline  
August 25th, 2016, 10:43 PM   #7
Member
 
Joined: Oct 2014
From: Mars

Posts: 30
Thanks: 0

Hi Brachydactyloid, thanks very much for your answer.

Do you know how I could solve this using a calculator? I can't read math notation

I tried doing: =((10/2)*(298/8 ))/(300/10) and I got 6.208?
Apple30 is offline  
August 26th, 2016, 05:04 AM   #8
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,112
Thanks: 1002

Quote:
Originally Posted by Apple30 View Post
Hi Brachydactyloid, thanks very much for your answer.

Do you know how I could solve this using a calculator? I can't read math notation

I tried doing: =((10/2)*(298/8 ))/(300/10) and I got 6.208?
Huh? This is like you've never heard of a frying pan and are trying to
fry eggs using a kettle

You're really asking Brach to conduct a classroom session.
Perhaps he will...

I suggest you do some work; go here:
https://www.khanacademy.org/math/pre...o-combinations
Denis is online now  
August 26th, 2016, 09:44 AM   #9
Senior Member
 
Joined: May 2016
From: USA

Posts: 1,310
Thanks: 550

Quote:
Originally Posted by Apple30 View Post
Hi all,

I have a very confusing problem to solve. No idea how to do it

Imagine there is a list of 300 items.
You must pick 10 items from that list. Other people must do the same.

Problem 1: What are the chances that any 2 items out of the 10 that you picked will match the same quantity (2) of another person?

and

Problem 2: What are the chances that any 2 or more items out of the 10 that you picked will match the same quantity of another person?


Thanks in advance,
A.
It is easiest to think about this if you think about matching a single item in the two lists. You have a list of 10. The other person has a list of 10, one item of which matches an item on your list. How many ways can that happen. TEN. The first item could be matched, or the second, or the third ....

Now 10 of the items are on your list, meaning 290 are NOT on your list. She must pick her 9 remaining items from those 290.

So the number of ways that exactly one item can be matched is

$\dbinom{10}{1} * \dbinom{290}{10 - 1}.$

If m is the number matched, the logic says that the number of ways is

$\dbinom{10}{m} * \dbinom{290}{10 - m}.$

Follow that?

So $\dbinom{10}{2} * \dbinom{290}{10 - 2} = \dfrac{10!}{2! * 8!} * \dfrac{290!}{8! * 282!} =$

$\dfrac{10 * 9}{2} * \dfrac{290 * 289 * 2888 * 287 * 286 * 285 * 284 * 283}{8 * 7 * 6 * 5 * 4 * 3 * 2} =$

$45 * 1,125,551,498,274,060 = 50,649,817,422,332,700.$

Now how many distinct lists can the other person choose?

$\dbinom{300}{10} = \dfrac{300!}{10! * 290!} =$

$1,398,320,233,241,701,770.$

So the probability of matching two items is $\dfrac{50,649,817,422,332,700}{1,398,320,233,241, 701,770} \approx 3.62\%.$

I'll let you complete the problem.

Last edited by JeffM1; August 26th, 2016 at 09:47 AM.
JeffM1 is offline  
August 26th, 2016, 09:50 AM   #10
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,112
Thanks: 1002

Jeff mon tabarnak, doesn't matter what we explain to poor Apple,
he is completely unfamiliar with combinations and stuff like nCm.
Denis is online now  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
combinations, number



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Max number of combinations Piggy Elementary Math 4 July 8th, 2016 03:42 PM
Total number of combinations crone Advanced Statistics 5 July 26th, 2013 01:01 PM
Number of Combinations? Pal Algebra 2 July 15th, 2013 06:55 PM
Combinations : regarding factors of number rnck Advanced Statistics 4 June 21st, 2013 04:26 PM
Number of combinations [solved] unnamed1 Applied Math 2 July 22nd, 2010 05:53 AM





Copyright © 2019 My Math Forum. All rights reserved.