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October 21st, 2015, 03:06 PM  #1 
Newbie Joined: Oct 2015 From: Finland Posts: 4 Thanks: 0  Need fast help for a hard equation
Hello! I just registered to this site to find a solution for the following problem: "There are 25 students. 10 Greek, 8 Finns, 2 Russians and 5 Swedes. The students study in groups. A group always consists of one or more students. If a group has at least 2 students that are of same nationality, then the group has to have at least one student who is different nationality. How many different ways are there to divide the 25 students to groups?" I would appreciate fast help! This equation is killing me as I am no maths expert... 
October 21st, 2015, 03:46 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
This looks hard! Without the restriction on ethnicity, the answer would be C(25), the 25th Catalan number. The most straightforward approach seems to be taking that as the starting point and subtracting off those groups which are disallowed. But that's not entirely straightforward since you can have instances with two or more illegal monoethnic groups and you don't want to doublecount these. 
October 23rd, 2015, 10:45 AM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 664 Thanks: 87 
Wikipedia says the 25th Catalan number is 1,289,904,147,324 which is 13 digits. Without any restriction on how many groups to make, I don't know how to solve it quickly. You could break it down into every amount of groups from 1 group of 25 to 25 groups of 1 (a group of 1 sounds strange but the problem says "one or more students," not two or more). For some amounts of groups the number of possibilities can be done quickly but for other amounts of groups it is hard. Here's a start: 1 group of 25: 1 way 24 groups (23 groups of 1 and 1 group of 2): 6 ways (the group of 2 must have two different nationalities so it's 4 C 2 = 6). 25 groups of 1: 1 way For 23 groups you would have to break it into 2 parts: 22 groups of 1 and 1 group of 3 21 groups of 1 and 2 groups of 2 For 10 groups you would have to break it into many parts (I don't know how many). What's the source of the problem? What math course are you taking? 
October 24th, 2015, 03:35 PM  #4 
Newbie Joined: Oct 2015 From: Finland Posts: 4 Thanks: 0 
This was actually on a Finnish game show where they're taking phone calls and giving out some money for the people who solve certain puzzles. No idea what the answer was as this was only a part of the puzzle... 
October 25th, 2015, 08:02 AM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Greek + nonGreek: 10 * 15 ways Finn + (Russian or Swede): 8 * 7 ways Russian + Swede: 2*5 ways Total: 216 ways  
October 25th, 2015, 12:50 PM  #6 
Newbie Joined: Oct 2015 From: Finland Posts: 4 Thanks: 0 
This is impossible. Last edited by Darien Fawks; October 25th, 2015 at 12:53 PM. 
October 27th, 2015, 11:26 AM  #7 
Member Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 
I thought of a possible solution, but I'm still trying to work it out and see if it makes enough sense :/ this is a lot of work!

October 27th, 2015, 05:18 PM  #8 
Member Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 
Yeah, I'm not entirely sure if this works, but any input would be appreciated. Basically you can think of the question as you can't have a group of more than one student with only one nationality. What if you took $\displaystyle C_{25}  C_{10}C_8C_5C_2$? I was thinking, wouldn't $\displaystyle C_{10}$ be the number of groups that could be formed where all students were Greek. $\displaystyle C_{8}$ for the Finns, $\displaystyle C_{5}$ for the Swedes and $\displaystyle C_{2}$ So, $\displaystyle C_{10}C_8C_5C_2$ would be all possible combinations where groups only had students from one ethnic group. Thus making $\displaystyle C_{25}  C_{10}C_8C_5C_2$ the answer. Is this valid logic? 
October 27th, 2015, 05:36 PM  #9 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  No. What about when all the groups are valid except the Russians are paired off? You're not counting it in $C_{10}C_8C_5C_2$ but it's still invalid.

October 31st, 2015, 04:12 PM  #10 
Newbie Joined: Oct 2015 From: Finland Posts: 4 Thanks: 0 
I guess no one can solve this. 

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