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Discrete Math 15 Online
OpenStudy (vortish):

A club has 20 members. a) How many ways are there to choose four members of the club to serve on an executive committee? [Num] ways b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office? [Num] ways

Miracrown (miracrown):

For problems like this, you have to decide if you're working with "combinations" or "permutations." Hopefully those sound familiar?

OpenStudy (anonymous):

this is a permutation/combination question are you familiar with the concept?

OpenStudy (anonymous):

part 1 would be 20C4, which is 20!/((20-4)!*4! do the math.

Miracrown (miracrown):

A combination is what you'd use if you're choosing a group of members, but in no particular order. So if you're picking a group of people to be on your team for tag. Permutations are when you're choosing for a group and the order DOES matter. So for example: choosing people to be the different positions on a baseball team (pitcher, catcher, etc.) So for part a)...

Miracrown (miracrown):

@lucian_denvers41325 Please refrain from giving direct solutions, it hinders the learning process.

OpenStudy (vortish):

yea i forgot how to work them I came up 116280 for part one and now just trying to remember part two

OpenStudy (vortish):

one is not right because i forgot a step

Miracrown (miracrown):

So notice that each person has a specific job, right? so that tells us it's which type? combination or permutation? @vortish

OpenStudy (vortish):

i came out with 4845 for part one

OpenStudy (anonymous):

ok, for part 2 out of the 4 people youve chosen for the executive committee, you get t0 pick one for each position. for the first position, you have 4 choices of people. you chose one of the 4 people. for the second post you have 3 choices, cuz you already reserved one person for the 1st post, and so on. i guess that should explain what you should do.

OpenStudy (vortish):

its a permutation

Miracrown (miracrown):

Right, and I got the same for part a) 4845

Miracrown (miracrown):

So back to the permutation in part b). What you can do is figure out how many options you have for the first position, which is the president. How many different people could be chosen to be president?

OpenStudy (vortish):

you have four for the 1st postison three for the second two for the third and one for the last

Miracrown (miracrown):

Well, there are 20 people total, though, right? So any of those 20 people could be chosen to be the president. So we have 20 options that for position.

Miracrown (miracrown):

B) _20_ __ __ ___

Miracrown (miracrown):

Now after you choose the president... how many people are left to choose the vice pres from?

Miracrown (miracrown):

Remember - no person can hold more than one office.

OpenStudy (vortish):

19 for the second 18 for the third and 17 for the last

Miracrown (miracrown):

So we can't use whichever person was just chosen to be president.

Miracrown (miracrown):

And right!

Miracrown (miracrown):

Then you just multiply those. So like this: 20 * 19 * 18 * 17 = ?

OpenStudy (vortish):

116280

Miracrown (miracrown):

Yay! Good job! =]

OpenStudy (vortish):

k thanks

Miracrown (miracrown):

yw

OpenStudy (vortish):

can you help me with another one

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