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Speaking Volumes (Posted on 2004-11-10) Difficulty: 3 of 5
In a group of students, 50 speak English, 50 speak French and 50 speak Spanish. Some students speak more than one language. Prove it is possible to divide the students into 5 groups (not necessarily equal), so that in each group 10 speak English, 10 speak French and 10 speak Spanish.

See The Solution Submitted by Brian Smith    
Rating: 3.1667 (6 votes)

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Solution Solution (possibly too informal) | Comment 1 of 19

I don’t know if this could be considered an official/elegant proof, but here’s how I got there… The minimum number of students you could have is 50, where they are all trilingual. The maximum you could have is 150, where each is monolingual.

Now I could approach this two different ways. I could start with 50 students, and show that I can increase the number of students by 1 by "splitting" a trilingual person and continue from there. Or I could start with 150 students, and show that I can decrease the number of students by 1 by "merging" two monolingual people and continue from there.

What the heck, I’ll do both. If you get bored after the first, just don’t read the next =)

Ok, start with 50 students that are all trilingual. Clearly it is possible to divide these students into 5 groups so that in each there are 10 English, 10 French, and 10 Spanish speakers. Simply divide them into 5 groups of 10. Since the 10 in any group are trilingual, there will always be 10 of each language speaker.

Ok, what about when there are 51-150 students? Well, let’s start with the grouping arrangement of 50 students and "grow" from there. Take someone who is in Group A, and "split" them. In other words, replace one trilingual person with 1 bilingual and 1 monolingual (making sure the monolingual doesn’t speak a language the bilingual does). So for example, I could replace a trilingual person with 1 person that speaks French and English, and 1 person that just speaks Spanish. This splitting allows me to increase the number of people by one while keeping the 50-50-50 language criteria intact.

If I keep those two people in Group A when I remove the trilingual person, the division distribution stays the same. I still have 5 groups with 10 English, 10 French, and 10 Spanish speakers.

I can continue doing this, either splitting a trilingual person into a bilingual and a monolingual, or splitting a bilingual into two monolinguals. Each time I will increase the number of students by 1, I will keep the 50-50-50 language criteria intact, and the 5 groups will aways have 10-10-10 English, French, and Spanish speakers as long as the two post-split people stay in the same group as the pre-split person. I can do this all the way up until there are 150 monolingual students and I have 5 groups of 30 monolinguals, where each group has 10 English, 10 French, and 10 Spanish speakers. The other proof is to start with 150 monolingual students. I can group those into 5 groups of 30 monolinguals where each group has 10 English, 10 French, and 10 Spanish speakers. I can pick two people (who don’t speak the same language) in Group A, for example, and "merge" them into one person that is bilingual in those two languages.

In doing this merging, I am decreasing the total number students by 1, but I am retaining the overall 50-50-50 criteria, as well as keeping the 10-10-10 criteria in each group. I can continue doing this, either merging two monolinguals who speak different languages into one bilingual, or merging one bilingual and one monolingual (where the monolingual doesn’t speak a language that the bilingual does) into a trilingual. I can do this all the way down until there are 50 trilingual students and I have 5 groups of 10 trilingual students.

I like the first proof better.  In the second proof I should really show that there will always be two monolinguals who speak different languages, or that there will always be a monolingual who doesn't speak a language that the bilingual speaks, in order to prove that I can always create a legal bilingual or trilingual.  So that's why I like the first proof better.


  Posted by nikki on 2004-11-10 10:45:34
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