Question 1082574: Five people from a class of 32 are to be chosen.
How many ways can they be chosen to serve on the student activities committee? Show all work
How many ways can they be chosen to be president, Vice President, secretary, treasurer, and publicity manager? Show all work
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Part 1:
Question: Five people from a class of 32 are to be chosen.
How many ways can they be chosen to serve on the student activities committee? Show all work
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Answer: 201376
Note: written with commas, the answer is 201,376 (a little over 201 thousand)
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Explanation:
See part 2 below. You'll follow the same steps you see in part 2 to get the value of 24165120; however, there are more steps to take. I won't repeat the steps because it's literally going to be verbatim. So it makes sense to save time/space here.
Let's say we had people labeled as "person1","person2", etc etc all the way up to "person32". I'm assigning numeric values to keep things uniform and simple.
Now let's say we pick a random group of 5 people, say this group: {person8, person 4, person9, person12, person27}
If there are no ranks or distinctions of positions, then we know that order does NOT matter. With any committee this is true (with the exception of people like a chairperson but ignore those wrinkles). Since order does NOT matter, we run into the problem that the previous result (in part 2) 24165120 is way too high. It overcounts big time.
A group like {person8, person 4, person9, person12, person27} is the same as {person8, person 27, person9, person12, person4}. All I did was swap person4 and person27. It's still the same committee. Each member has their own equal say in whatever decision happens.
So effectively in that example, {person8, person 4, person9, person12, person27} = {person8, person 27, person9, person12, person4} meaning we're at least double counting, if not worse. Turns out the overcount is by a factor of 120. How did I find this out? By computing 5! = 5*4*3*2*1 = 120 where that "5" in "5!" refers to the number of slots to fill.
There are 120 different ways to arrange 5 people in 5 slots where order matters. If order doesn't matter, then those 120 outcomes are all the same group.
So we have 24165120 different permutations (result from part 2) if order mattered, but order doesn't matter. To fix this overcounting error, we divide 24165120 over 120
24165120/120 = 201376
which is the final answer.
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If you wish to use a formula, then use the combination formula
 take note of the paired "27!" terms
 which cancel out and go away
yielding the same answer as before.
Note: there is a connection between the combination formula and the permutation formula, namely,
which basically says "whatever the permutation result is, divide by r factorial and you'll get the combination result". In this case, the permutation result was 24165120 and r = 5 so r! = 5! = 120 was that factor I used to help correct for overcounting.
If you choose to go this route, then you'd say
showing yet another way to get the answer. Though to be fair, this isn't completely unique since it's very similar to the previous section above.
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Part 2:
Question: Five people from a class of 32 are to be chosen.
How many ways can they be chosen to be president, Vice President, secretary, treasurer, and publicity manager? Show all work
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Answer: 24165120
Note: written with commas, the answer is 24,165,120 (a little over 24 million)
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Explanation:
Start off by drawing five empty boxes that are side by side in a single row. This drawing is completely optional and only if you need a visual guide. Each box represents a slot to fill with a name from a person selected.
Let's label the five slots: slot A, slot B, slot C, slot D, slot E
Slot A has 32 choices from the current pool of 32 people.
Once a selection has been made, that person cannot be picked again. So we have 32-1 = 31 choices for slot B.
Slot C has 30 choices
Slot D has 29 choices
Slot E has 28 choices
Notice how I'm counting down from 32 onto 28. I stop once I've filled all the boxes. In other words, I stop counting down once I've made 5 selections.
Those values (32,31,30,29,28) are then multiplied out to get: 32*31*30*29*28 = 24165120
So there are 24165120 different permutations.
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An alternative is to use the permutation formula to get
 take note of the paired "27!" terms
 which cancel out and go away
and we get the same answer
Note: order matters in this case. This is because the various positions are different from one another.
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