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?

Guest Mar 18, 2015

#1**+10 **

Best Answer

Notice that if you had 11 friends and you choose any four of them, there would be

C (11,4) = 330 different combinations.....not enough!!!

And 12 friends would produce

C(12, 4) = 495 different combinations

So you would need at least 12

CPhill
Mar 19, 2015

#2**+5 **

I used formula to get a vague starting point of 10 but then I would do it by trial and error after that the same as CPhill did.

$$\\^nC_4\ge 365\\\\

\frac{n!}{4!(n-4)!}\ge 365\\\\

\frac{(n-4)!*(n-3)*(n-2)*(n-1)*n}{24(n-4)!}\ge 365\\\\

(n-3)*(n-2)*(n-1)*n\ge 365*24\\\\

n^4.....\ge 8760\\\\$$

so $${\sqrt[{{\mathtt{{\mathtt{4}}}}}]{{\mathtt{8\,760}}}} = {\mathtt{9.674\: \!444\: \!255\: \!582\: \!887\: \!1}}$$

My first 'guess' will be 10. Now I will jut try 1 by 1 till I hone in on the right answer. :)

Melody
Mar 19, 2015

#3**+5 **

Melody actually didn't quite carry her calculations far enough.

We have

(n-3)(n-2)(n-1)n ≥ 8760

n^{4} - 6n^{3} + 11n^{2} - 6n ≥ 8760

Using WolframAlpha ...we get the following (somewhat messy) solution....

n ≥ (3/2) + (1/2)* sqrt(5 + 4√8761) =

n ≥ about 11.239

Which confirms that we need more than 11.... (which means 12)

Her way is actually superior to my "guess and check" approach....!!!

But.....as she always says..."All roads lead to Rome"

CPhill
Mar 19, 2015