To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which selection can be made?

• First consider the hiring of the 3 programmers. You have 3 slots to fill.
  
  When fillling the first slot you have 6 candidates to choose from. For the 2nd slot you can
  choose from 5 remaining candidates, then 4 for the 3rd slot.
  
  This gives 6 x 5 x 4 ways to choose.
  
  However, this includes variations in order (i.e. ABC, BCA, etc) and for this sort of choosing
  order isn't important. You must divide by the number of ways that 3 people can be ordered to
  account for this. This is easy enough, the number of ways the 3 people can be ordered is
  3! = 3 x 2 x 1. So the first part of the problem is 6 x 5 x 4/(3 x 2 x 1) = 20
  
  
  The 2nd part works the same way. You have 2 slots to fill. 4 candidates for the first slot
  and 3 for the 2nd. So you start out with 4 x 3. But you have to divide by the number of ways
  that 2 people can be ordered is 2! = 2 (AB, BA). You have 4 x 3/2 = 6
  
  So you have 20 ways to pick the programmers and 6 ways to pick the managers.
  
  Since these are independent sets, the number of ways that this selections combine is the
  product of the two.
  
  20 x 6 = 120
• 12 years agoHelpfull: Yes(4) No(0)
• c(6,3)*c(4,2)=20*6=120
• 12 years agoHelpfull: Yes(0) No(1)