Question 1194824: A man has a total of n friends; he wants to invite three of them for dinner every day during the first 4 months of the year 2023. Calculate the minimum number of friends , n, he must have if he does not want to invite the same combination of three friends more than once?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
If we look at a calendar of 2023, then we'd find that:- January has 31 days
- February has 28 days (non leapyear)
- March has 31 days
- April has 30 days
There are a total of 31+28+31+30 = 120 days in the first four months.
This must mean that there must be 120 or more combinations of three friends to invite friends over.
Recall that the nCr combinations can be found in Pascal's Triangle.
Go to this page here (or wherever your favorite resource is to look up Pascal's Triangle)
https://www.mathsisfun.com/pascals-triangle.html
Scroll down until you reach the section titled "The First 15 Lines".
You'll find a pretty massive triangle.
Looking through that triangle shows that 120 is buried inside it.
It's found on the row that has 1, 10, 45, 120, ...
The second item of this row is the value of n we're after.
You should find that if n = 10 and r = 3, then nCr = 10C3 = 120
Scratch work:
nCr = (n!)/(r!*(n-r)!)
10C3 = (10!)/(3!*(10-3)!)
10C3 = (10*9*8*7!)/(3!*7!)
10C3 = (10*9*8)/(3!)
10C3 = (10*9*8)/(3*2*1)
10C3 = 720/6
10C3 = 120
If the man has n = 10 friends, then he has 120 different trios of friends to invite over.
This ensures that he has a different trio on every night for the first four months of 2023.
Keep in mind that repeated friends can be allowed (i.e. the same friend invited on multiple nights), just not repeated trios.
For instance, let's say the man has ten friends with code names A,B,C,D,E,F,G,H,I,J
One group to invite over is {A,B,C} for night one, then on night #2 could be the group {A,C,D}
The group {A,C,B} is a repeat of the first trio, so it's not allowed.
Order doesn't matter within any group.
If n > 10, then he'll have more possible trios.
Therefore, n = 10 is the smallest possible value of n.
Answer: 10 friends.
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