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soccerfool14

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  1. (a) Suppose a club has 20 members. Find the number of ways of choosing anexecutive committee containing a president, vice-president, treasurer, andsecretary from the club members, assuming that Alice and Bob refuse toserve on the same committee.




Can someone walk me through the setup of this probability problem? Thanks.
 
  1. (a) Suppose a club has 20 members. Find the number of ways of choosing anexecutive committee containing a president, vice-president, treasurer, andsecretary from the club members, assuming that Alice and Bob refuse toserve on the same committee.




Can someone walk me through the setup of this probability problem? Thanks.
This is actually a counting problem, not a probability problem.

There probably are several ways to solve this. Here is the one that I find most intuitive although not the quickest.

Let X be the set of executive committees that exclude Alice and Bob.

Let Y be the set of executive committees that include Alice but exclude Bob.

Let Z be the set of executive committees that exclude Alice and include Bob.

What is \(\displaystyle n\left(X \bigcup Y \bigcup Z\right)\), where n(S) means the number in the set S.
 
  1. (a) Suppose a club has 20 members. Find the number of ways of choosing an executive committee containing a president, vice-president, treasurer, and secretary from the club members, assuming that Alice and Bob refuse to serve on the same committee.

Can someone walk me through the setup of this probability problem? Thanks.
The question is ambiguous. It says "select a committee" suggesting that it doesn't matter who fills which office, since all are members of the committee. In that case, figure it out using Combinations.

On the other hand, it does list specific officers. If you interpret that to mean that it does matter which is which, then figure it out using Permutations.

Which do you think it is? Are you having trouble with the Alice/Bob conflict? We need to see your work to help where you are getting stuck.
 
Total number of committees including those with alice and bob = (20 P 4), right? (20 total people, 4 on a committee, order doesn't matter, so it's a permutation?)
And that means that the number of committees that include alice and bob would be (18 P 2)? (18 other people to choose from, with 2 spots, again order doesn't matter so permutation again?)
So answer would be (20 P 4) - (18 P 2)?
That's how I interpret the question, obviously I'm interpreting it wrong though, because the solution shows the answer to be (20P4)- 4!(18P2). Where does the 4! come from..?
 
Total number of committees including those with alice and bob = (20 P 4), right? (20 total people, 4 on a committee, order doesn't matter, so it's a permutation?) NO - use P because order DOES matter
And that means that the number of committees that include alice and bob would be (18 P 2)? (18 other people to choose from, with 2 spots, again order doesn't matter so permutation again?)
So answer would be (20 P 4) - (18 P 2)?
That's how I interpret the question, obviously I'm interpreting it wrong though, because the solution shows the answer to be (20P4)- 4!(18P2). Where does the 4! come from..?
Use Combinations when order doesn't matter. If the book answer is in terms Permutations, that means order does matter.

The 4! should be because once you have selected Alice, Bob, and 2 others, there are 4! ways to assign those four individuals to the offices. I am not fully satisfied with that result, because if you use P then the 2 random people are already ordered between themselves. I would prefer 4!(18C2) for the permutations with both Alice and Bob present.

Not only is the question ambiguous - so is the answer.
 
Total number of committees including those with alice and bob = (20 P 4), right? (20 total people, 4 on a committee, order doesn't matter, so it's a permutation?)
And that means that the number of committees that include alice and bob would be (18 P 2)? (18 other people to choose from, with 2 spots, again order doesn't matter so permutation again?)
So answer would be (20 P 4) - (18 P 2)?
That's how I interpret the question, obviously I'm interpreting it wrong though, because the solution shows the answer to be (20P4)- 4!(18P2). Where does the 4! come from..?

As Phil said, order does matter. But the given answer is still wrong.

If should be \(\displaystyle _{20}P_{4}-4!~_{18}C_{2}\)

To explain: the first number is the total possible committees.
The \(\displaystyle _{18}C_{2}\) is the number of sets of four that contain both Alice and Bob.
The \(\displaystyle 4!\) accounts for ordering those sets of four.
 
Hello, soccerfool14!

Total number of committees including those with Alice and Bob = (20 P 4), right?
(20 total people, 4 on a committee. .Order does matter, so it's a permutation.)

And that means that the number of committees that include Alice and Bob would be (18 P 2).
(18 other people to choose from with 2 spots. .Again order does matter, so permutation again.)

So answer would be (20 P 4) - (18 P 2)?

That's how I interpret the question, obviously I'm interpreting it wrong though.
Because the solution shows the answer to be (20 P 4)- 4!(18 P 2). .I don't agree!
Where does the 4! come from?

I agree with the first part.

We select 4 people from the entire 20 available people
. . and assign them to specific positions.
There are: .\(\displaystyle _{20}P_4\) possible executive committees.

How many committees contain both Alice and Bob?
Place Alice and Bob on the committee.
Then choose 2 of the other 18 people.
There are: .\(\displaystyle _{18}C_2\) ways.

We have a set of four people: .\(\displaystyle \{A,B,X,Y\}\)
Now we assign them to specific positions.
There are \(\displaystyle 4!\) possible assignments.

Hence, there are: .\(\displaystyle 4!(_{18}C_2)\) committees with both Alice and Bob.

Therefore, the answer is: .\(\displaystyle \boxed{_{20}P_4 - 4!(_{18}C_2)}\)


We have: .\(\displaystyle (20\cdot19\cdot18\cdot17) - (24)\left(\frac{18\cdot17}{2\cdot1}\right) \)

. . . . . . \(\displaystyle =\;116,\!280 - 3,\!672 \;=\;\boxed{112,\!608}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

There are three cases to consider.

[1] Alice is on the committee; Bob is not.
. . .Choose 3 people from the other 18 people.
. . .There are \(\displaystyle _{18}C_3\) choices.
. . .There are \(\displaystyle 4!\) ways to assign their positions.
There are: .\(\displaystyle 4!(_{18}C_3)\) committees with Alice (only).

[2] Bob is on the committee; Alice is not.
. . .Choose 3 people from the other 18 people.
. . .There are \(\displaystyle _{18}C_3\) choices.
. . .There are \(\displaystyle 4!\) ways to assign their positions.
There are: .\(\displaystyle 4!(_{18}C_3)\) committees with Bob (only).

[3] Neither Alice nor Bob is on the committee.
. . .Choose 4 people from the other 18 people.
. . .There are \(\displaystyle _{18}C_4\) choices.
. . .There are \(\displaystyle 4!\) ways to assign their positions.
There are: .\(\displaystyle 4!(_{18}C_4)\) committees with neither Alice or Bob.


Answer: .\(\displaystyle 4!(_{18}C_3) + 4!(_{18}C_3) + 4!(_{18}C_4) \)

. . . . . . \(\displaystyle =\;19,\!584 + 19,\!584 + 73,\!440\)

. . . . . . \(\displaystyle =\; \boxed{112,\!608} \;\;\checkmark\)
 
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