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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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angle AOB = 132 and is also the sum of angles AOD and DOB. Hence 
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angle COD = 141 and is also the sum of angles COB and BOD. Hence 
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Now we add the left sides together and the right sides of equations 1 and 2 together to form a new equation. 

angle AOD + angle DOB + angle COB + angle DOB = 132 + 141       ---> 3 

We should also note that: 
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Therefore substituting angle AOD + angle DOB + angle COB in equation 3 by 180 and solving for angle DOB:
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