SOLUTION: 10 people are to be divided into 3 committees, in such a way that every committee must have at least one member. and no person can serve on all three committees. In how many ways c

Algebra ->  Permutations -> SOLUTION: 10 people are to be divided into 3 committees, in such a way that every committee must have at least one member. and no person can serve on all three committees. In how many ways c      Log On


   

Question 610605: 10 people are to be divided into 3 committees, in such a way that every committee must have at least one member. and no person can serve on all three committees. In how many ways can this be done.
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
There are 8 ways to divide into 3 committees: 

   1st        2nd        3rd
committee  committee  committee
 has this   has this   has this 
  many       many       many
 members    members    members
    1          1          8      C(10,1)·C(9,1)·C(8,8)/2 = 10·9·1/2 = 45 ways  
    1          2          7      C(10,1)·C(9,2)·C(7,7)=10·36·1 = 360 ways  
    1          3          6      C(10,1)·C(9,3)·C(6,6)=10·84·1 = 840 ways
    1          4          5      C(10,1)·C(9,4)·C(5,5)=10·126·1 = 1260 ways
    2          2          6      C(10,2)·C(8,2)·C(6,6)/2 = 45·28·1/2 = 630 ways
    2          3          5      C(10,2)·C(8,3)·C(5,5) = 45·56·1 = 560 ways
    2          4          4      C(10,2)·C(8,4)·C(4,4) = 45·70·1/2 = 1575 ways
    3          3          4      C(10,3)·C(7,3)·C(4,4) = 120·35·1/2 = 2100 ways

Note: In the four cases where there are two committees of the same size,
we must divide by 2 in order to avoid ordering those two committees.

Total = 45+360+840+1260+630+560+1575+2100 = 7370 ways.

Edwin