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Marina86 [1]
2 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]2 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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We can solve this problem by stating that the summation of momentums must be zero. That is, the momentum at one end of the pivot must be equal to the momentum at the other end.

Momentum 1 = Momentum 2

Since Momentum is the product of Force and Distance, and with this rule created, we can say that:

F1 * d1 = F2 * d2

Where,

F1 = 60 lbs

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Substituting to the equation to find for F2:

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F2 = 40 lbs

 

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