
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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August 25th, 2016, 05:09 AM  #1 
Member Joined: Oct 2014 From: Mars Posts: 35 Thanks: 0  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. 
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 nonmatching 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%. 
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 $\sum_{i=2}^{10}\binom{10}{i}\binom{290}{10i}$, the sum of all desirable matching values. The denominator is the same as before. The answer is $$\frac{\sum_{i=2}^{10}\binom{10}{i}\binom{290}{10i}}{\binom{300}{10}}$$ 
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  
August 25th, 2016, 11:23 AM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Quote:
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?????????????????  
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  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.

August 25th, 2016, 10:43 PM  #7 
Member Joined: Oct 2014 From: Mars Posts: 35 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? 
August 26th, 2016, 05:04 AM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Quote:
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...ocombinations  
August 26th, 2016, 09:44 AM  #9  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
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.  
August 26th, 2016, 09:50 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
Jeff mon tabarnak, doesn't matter what we explain to poor Apple, he is completely unfamiliar with combinations and stuff like nCm. 

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