SOLUTION: The neighborhood association elects a president, vice president, secretary and treasurer.How many sets of officers are possible if there are 16 people in the association?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The neighborhood association elects a president, vice president, secretary and treasurer.How many sets of officers are possible if there are 16 people in the association?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 167633: The neighborhood association elects a president, vice president, secretary and treasurer.How many sets of officers are possible if there are 16 people in the association?
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
you can make (16*15*14*13)/(1*2*3*4) possible combinations of 4 people out of 16 people.
-----
this is the same as %2816%21%29%2F%28%2812%21%29%2A%284%21%29%29
where 16 = number of people in universe to pick from (n),
and 4 = number of people you want in each set (x)
general formula is: %28n%21%29%2F%28%28%28n-x%29%21%29%2A%28x%21%29%29
-----
the number comes out to be 1820 possible sets of 4 out of 16 people.
-----
what is happening is easier to see if the universe is smaller.
say you wanted a set of 2 people out of a universe of 3 people.
n = 3
x = 2
n!/(1!*2!) = 3*2*1/1*2*1 = 3
there are 3 possible sets of 2 out of 3.
assuming the elements in the universe are abc, then the possible combinations are:
ab
ac
bc
-----
if you wanted 2 out of 4, then the number of possible sets would be
4! / (2!*2!) = 4*3*2*1/2*1*2*1 = 6
assuming elements in your universe are abcd, then
elements in each set would be:
ab
ac
ad
bc
bd
cd
-----
your problem is much more complex but solves using the same method.
let's say your universe was abcdefghijklmnop
then possible sets of 4 would be
abcd
abce
abcf
abcg
etc.,
all the way through to...
lmno
lmnp
mnop
-----