Question 1081075

a. 

How many different committees of {{{5}}} can be formed?
Order is not important in this situation, so the selection is a combination of
{{{15}}} people chosen {{{5}}} at a time.

{{{C(15, 5) =15!/(15-5)!5!=15!/(10!5!)=(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/((10*9*8*7*6*5*4*3*2*1)(5*4*3*2*1))=3003}}}

There are {{{3003}}} different ways to form the committees of {{{5}}}.


b.
Order has to be considered in this situation because each committee
member has a different responsibility.

{{{P(15, 5) =15!/(15-5)!=15!/(10!5!)=(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(10*9*8*7*6*5*4*3*2*1)=360360}}}
There are {{{360360}}} possible committees.


c.
Order is not important. There are three questions to consider.
How many ways can {{{2}}} boys be chosen from {{{7}}}?
How many ways can {{{3}}} girls be chosen from {{{8}}}?

Then, how many ways can {{{2}}} boys and {{{3}}} girls be chosen together?
Since the events are independent, the answer is the product of the combinations 
{{{C(7, 2)}}} and {{{C(8, 3)}}}.

{{{C(7, 2)*C(8, 3)=(7!/(7-2)!2!)*(8!/(8-3)!3!) =8!/(5!3!)=21*56=1176}}}

There are {{{1176}}} possible committees.