Question 1007417: An environmental organization has 19 members. Each member will be placed on exactly 1 of 4 teams. Each team will work on a different issue. The first team has 3 members, the second has 5, the third has 7, and the fourth has 4. In how many ways can these teams be formed?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe your solution is going to be:
19c3 * 16c5 * 11c7 * 4c4
ncx is the combination formula of n! / (x! * (n-x)!)
what happens, i believe, is this:
your first team is composed of 3 people taken from 19.
that gets you 19c3 = 969 possible teams.
for each of these teams, there are 16 members left that can be used to form the second team.
the second tesm is composed of 5 people taken from 16.
that gets you 16c5 = 4368 possible teams for EACH of the 969 first teams.
for each of these second teams, there are 11 members left that can be used to form the third team.
the third team is composed of 7 people taken from 11.
that gets you 11c7 = 330 possible teams for EACH of the 4368 second teams.
for each of the third teams, there are 4 members left that can be used to form the fourth team.
the fourth team is composed of 4 people taken from 4.
that gets you 4c4 = 1 possible team for EACH of the 330 third teams.
you get:
19c3 * 16c5 * 11c7 * 4c4 becomes:
969 * 4368 * 330 * 1 = 1396755360 possible teams that can be formed.
does that make sense?
let's see with a little experiment.
suppose 4 teams are formed.
3 teams are composed of 2 members and the last team is composed of 1 member.
that would be 7 members forming 4 teams.
the team members are abcdefg.
the first team is formed as 7c2 = 21 possible teams.
they are:
ab
ac
ad
ae
af
ag
bc
bd
be
bf
bg
cd
ce
cf
cg
de
df
dg
ef
eg
fg
now let's look at just one of these teams.
let's look at the team of ab.
when the team of ab is chosen, we have 5 members left that can form the second team of 2 members.
that second team can be formed from cdefg
the number of possible teams is 5c2 = 10.
the members of the second team are:
cd
ce
cf
cg
de
df
dg
ef
eg
fg
now we'll look at the possible third teams that can be formed when just the first of the second teams has been formed.
ab are the chosen team from the first team.
cd are the chosen team from the second team.
when ab and cd have been chosen for the first and second teams, the remaining members are efg
the third team can be formed in 3c2 possible ways.
those ways are:
ef
eg
fg
by now you should get the idea that the formula is good.
you get 7c2 * 5c2 * 3c2 * 1c1
what makes combinations and permutations so difficult is that it is very hard to prove the formulas are correct.
the way that i've found is to simplify the problem to the point where you can show that the formulas work.
i think i've done that here.
i'm reasonably confident the solution is correct.
if you know what the solution is supposed to be, you can check against it to see if you did it right.
o0therwise, there's no way of knowing unless you do something similar to what i've done above.
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