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In Maricopa County, 5 persons are to be elected to the Board of Supervisors. If 8 persons are candidates, how many different arrangements are possible?

6,720

336

56

User Bobdiaes
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1 Answer

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The question asks about "arrangements". That seems to mean that
the position of the candidate's name on the list matters.

The first name can be any one of the 8. For each one . . .
The second name can be any one of the remaining 7. For each one . . .
The third name can be any one of the remaining 6. For each one . . .
The fourth name can be any one of the remaining 5. For each one . . .
The fifth name can be any one of the remaining 4.

The total number of possible 'arrangements' = (8 x 7 x 6 x 5 x 4) = 6,720 .
(That's 8! / 3! .)

But for this type of problem, the order usually doesn't matter ... it's all
the same thing whether 'Jones' is 1st, 3rd, or 5th on the list etc.
Then the question is really: How many different groups of 5 can be
nominated with 8 candidates ?

Well, how many different ways can you arrange 5 names ?
Using the same type of reasoning as above, 5 names can be
lined up in (5 x 4 x 3 x 2 x 1) = 120 different orders.

So, while it's true that you can make 6,720 different 5-name line-ups
with 8 names, every possible group appears on that huge list shuffled
in 120 different orders.

If the order doesn't matter, then there are only (6,720 / 120) = 56 different
groups of 5 names, although they appear in every possible shuffle on the
big list.

User Naveen Babu
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