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.

Let the following letters symbolize the language speakers:

Then in the entire group of people,
 
50 = e + ef + es + efs
50 = f + ef + fs + efs
50 = s + es + fs + efs
 
Because all three groupings sum to 50, it is always possible to split the group into five groups, in each of which 10 will speak English, 10  French and 10 Spanish, by first going after all
who speak all three languages, then those who speak pairs
of languages, and finally those who speak only  one language. Every time you take someone from a pool of two or three languages, there will be exactly the right number  of single language speakers in each case to complete  the five groups, as needed, because of those three equations to 50.    
    
Case 1: Everyone speaks only one language.

e=50 f=50 s=50
   The groups are:
   10e + 10f + 10s
   10e + 10f + 10s
   10e + 10f + 10s
   10e + 10f + 10s
   10e + 10f + 10s
 
Case 2: There is just one pair of languages  that one or more people speak. No one  speaks all three languages.

fs= 1 e= 50 f= 49 s= 49
    The groups are: 
    fs + 10e + 9f + 9s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s    
    
fs= 2 e= 50 f= 48 s= 48
    Groups:
    2fs + 10e + 8f + 8s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
       
 etc...
 
fs= 50 e= 50
    Groups:
    10fs + 10e
    10fs + 10e
    10fs + 10e
    10fs + 10e
    10fs + 10e
 
Case 3: There are two pairs of languages  that one or more people speak. No one  speaks all three.  

es= 1 fs= 1 e= 49 f= 49 s= 48
     Groups:
     es + fs + 9e + 9f + 8s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
    
es= 1 fs= 2 e= 49 f= 48 s= 47
     Groups:
     es + 2fs + 9e + 8f + 7s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
    
etc....    
 
es= 49 fs= 1 e= 1 f= 49
     Groups:
     9es + fs + e + 9f
     10es + 10f
     10es + 10f
     10es + 10f
     10es + 10f
 
Case 4: There are three pairs of  languages that one or more people  speak. No one speaks all three.
 
ef= 1 es= 1 fs= 1 e= 48 f= 48 s= 48
      Groups:
      ef + es + fs + 8e + 8f + 8s
      10e + 10f + 10s
      10e + 10f + 10s
      10e + 10f + 10s
      10e + 10f + 10s 
  
ef= 1 es= 1 fs= 2 e= 48 f= 47 s= 47
      Groups:
      ef + es + 2fs + 8e + 7f + 7s
      10e + 10f + 10s
      10e + 10f + 10s
      10e + 10f + 10s
      10e + 10f + 10s

etc...
 
ef= 49 es= 1 fs= 1 s= 48
      Groups:
      9ef + es + fs + 8s 
      10ef + 10s
      10ef + 10s
      10ef = 10s
      10ef + 10s
 
Case 5: At least one person speaks all  three languages. No one speaks just two.
 
efs= 1 e= 49 f= 49 s= 49
     Groups:
     efs + 9e + 9f + 9s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f = 10s
 
efs= 2 e= 48 f= 48 s= 48
     Groups:
     2efs + 8e + 8f + 8s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f + 10s
     10e = 10f = 10s
 
etc...
 
efs = 50
     Groups:
     10efs
     10efs
     10efs
     10efs
     10efs
 
Case 6: There is one pair of languages that one or more people speak. At least one person speaks all  three.
 
efs= 1 fs= 1 e= 49 f= 48 s= 48
     Groups:
     efs + fs + 9e + 8f + 8s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f = 10s
     10e + 10f + 10s
    
efs= 1 fs= 2 e= 49 f= 47 s= 47
     Groups:
     efs + 2fs + 9e + 7f + 7s
     10e + 10f + 10s
     10e + 10f + 10s
     10e + 10f = 10s
     10e + 10f + 10s
    
etc...
 
efs= 49 fs= 1 e= 1
     Groups:
     9efs + fs + e
     10efs
     10efs
     10efs
     10efs

Case 7: There are two pairs of languages that one or more people speak. At least one person speaks all three.

efs= 1 es= 1 fs= 1 e= 48 f= 48 s= 47
    Groups:
    efs + es + fs + 8e + 8f + 7s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s

efs= 1 es= 1 fs= 2 e= 48 f= 47 s= 46
    Groups:
    efs + es + 2fs + 8e + 7f + 6s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
   
etc...

efs= 48 ef= 1 es= 1 f= 1 s= 1
    Groups:
    8efs + ef + es + f + s
    10efs
    10efs
    10efs
    10efs

Case 8: There is at least one person who only speaks one of three possible pairs of languages. There is at least one person who speaks all three.

efs= 1 ef= 1 es= 1 fs= 1 e= 47 f= 47 s= 47
    Groups:
    efs + ef + es + fs + 7e + 7f + 7s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s

efs= 1 ef= 1 es= 1 fs= 2 e= 47 f= 46 s= 46
    Groups:
    efs + ef + es + 2fs + 7e + 6f + 6s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
    10e + 10f + 10s
   
etc...

efs= 48 ef= 1 es= 1 fs= 1
    Groups:
    8efs + ef + es + fs
    10efs
    10efs
    10efs
    10efs

Case 9: Everybody speaks all three languages.
efs= 50

    Groups:
    10efs
    10efs
    10efs
    10efs
    10efs

 

 

Posted by Penny on 2004-11-13 13:34:26