SOLUTION: in how many ways can 9 people petitioned into committees containing 4,3 and 2 members respectively? a- 9!/9!3!2! b- 9 c- 9! 4! 3! 2! d- 4! 3! 2! e- 3. 4!3!2!

Algebra ->  Permutations -> SOLUTION: in how many ways can 9 people petitioned into committees containing 4,3 and 2 members respectively? a- 9!/9!3!2! b- 9 c- 9! 4! 3! 2! d- 4! 3! 2! e- 3. 4!3!2!      Log On


   

Question 912042: in how many ways can 9 people petitioned into committees containing 4,3 and 2 members respectively?
a- 9!/9!3!2!
b- 9
c- 9! 4! 3! 2!
d- 4! 3! 2!
e- 3. 4!3!2!
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
We choose the 4 in 9C4 ways.  That's %289%2A8%2A7%2A6%29%2F%284%2A3%2A2%2A1%29
We choose the 3 in 5C3 ways.  That's %285%2A4%2A3%29%2F%283%2A2%2A1%29
We choose the 2 in 2C2 ways.  That's %282%2A1%29%2F%282%2A1%29

Multiplying all those together, that's



That's 9%21%2F%284%213%212%21%29

That's not listed, but I'm guessing choice c was supposed to be 9!/(4!3!2!).

Edwin