SOLUTION: A COMMITTEE OF 4 IS TO BE CHOSEN FROM 4 LAWYERS AND 3 ENGINERERS, FIND IN HOW MANY WAYS THE COMMITTEE CAN BE CHOSEN. IN HOW MANY WAYS THE COMMITTEE CAN BE EQUALLY REPRESENTED BY

Algebra ->  Permutations -> SOLUTION: A COMMITTEE OF 4 IS TO BE CHOSEN FROM 4 LAWYERS AND 3 ENGINERERS, FIND IN HOW MANY WAYS THE COMMITTEE CAN BE CHOSEN. IN HOW MANY WAYS THE COMMITTEE CAN BE EQUALLY REPRESENTED BY      Log On


   

Question 1201241: A COMMITTEE OF 4 IS TO BE CHOSEN FROM 4 LAWYERS AND 3 ENGINERERS, FIND
IN HOW MANY WAYS THE COMMITTEE CAN BE CHOSEN.
IN HOW MANY WAYS THE COMMITTEE CAN BE EQUALLY REPRESENTED BY THE TWO PROFESSIONS
Found 2 solutions by Glaviolette, ikleyn:
Answer by Glaviolette(140) About Me  (Show Source):
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
A COMMITTEE OF 4 IS TO BE CHOSEN FROM 4 LAWYERS AND 3 ENGINERERS, FIND
IN HOW MANY WAYS THE COMMITTEE CAN BE CHOSEN.
IN HOW MANY WAYS THE COMMITTEE CAN BE EQUALLY REPRESENTED BY THE TWO PROFESSIONS
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            Ignore the solution by  @Glaviolette:  it is totally wrong.
            I came to bring a correct solution. 


(1)  In  C%5B7%5D%5E4 = 7%21%2F%284%21%2A3%21%29 = %287%2A6%2A5%29%2F%281%2A2%2A3%29 = 35  different ways.


(2)  In  C%5B4%5D%5E2.C%5B3%5D%5E2 = 6*3 = 18 different ways.

Solved.