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I have friends. Every night of the 365-day year I invite 4 of them to dinner. What is the smallest could be such that it is still possible for me to make these invitations without ever inviting the same group of four friends?

rohitaop  May 18, 2018
edited by rohitaop  May 18, 2018
 #1
avatar+87604 
+1

We need to solve this

 

      n!  

________  =   365

(n -4)!* 4!

 

n!

_____   =   365 * 4!

(n - 4)!

 

n * ( n - 1) (n -2) (n - 3)  =  365 * 24

 

n^4 - 6 n^3 + 11 n^2 - 6n  = 8760

 

n^4  -6n^3 + 11n^2  - 6n - 8760   =   0

 

Solving this for n produces the positive solution  of  ≈ 11.239

 

So....we need 12 friends  to assure that the same 4 are not  invited

 

Proof

 

C(12,4)   = 495  different groups

C(11,4)  = 330    different groups  [ not enough ]

 

 

cool cool cool

CPhill  May 18, 2018

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