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Marina86 [1]
3 years ago
14

A 20-person fraternity is making their weekend plans around 5 different parties in IV. Each brother will attend exactly one of t

he parties. Fraternity rules state that the fraternity must have a representative at each party. Assume that only the number of brothers that show up at a party is important, i.e. the brothers are indistinguishable. (a) How many different arrangements of plans are possible
Mathematics
1 answer:
skad [1K]3 years ago
3 0

Answer:

3876

Step-by-step explanation:

Given the following :

Fraternity members = 20

They are to attend 5 different parties in groups of 4

Meaning a group = 4 persons

Atleast one brother will attend exactly one of the parties. (the brothers are indistinguishable).

Then, exactly one brother at a party (20 - 1) = 19, since they are indistinguishable.

Group members are in 4's

19C4

From: nCr = n! /(n-r)! r!

19C4 = 19! / (19 - 4)! 4!

= 19! / 15! 4!

= (19 * 18 * 17 * 16) / (4 * 3 * 2)

= 93024 / 24

= 3876

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Answer:

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Step-by-step explanation:

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Problems of normal distributions can be solved using the z-score formula.

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In this question:

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4 0
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