combinations
Huỳnh Anh Phương
30/09/2018 at 11:13
There are: 3*8*4=96 combinations that Billy can make. Huỳnh Anh Phương selected this answer.

FA KAKALOTS 09/02/2018 at 21:43
Let Cn be the answer for n points. We have C1 = p, C2 = p(p1) and C3=p(p1)(p2).
For n + 1 points, if A1 and An have differenr colors, then A1, ..., An can be colored in Cn ways, while An+1 can be colored in p  2 ways. If A1 and An have the same color, then A1, ..., An can be colored in Cn1 ways and An+1 can be colored in p  1 ways. So Cn+1 = (p2)Cn +(p1)Cn1 with n > 2 (*)
(*) can be written as:
Cn+1 + Cn = (p  1) (Cn + Cn1)
=> Cn+1 + Cn = (p  1)n2 (C3 + C2) = p(p  1)n.
By induction we infer that Cn = (p  1)n + (1)n (p  1)

london 11/04/2017 at 08:43
Let C_{n} be the answer for n points. We have C_{1} = p, C_{2} = p(p1) and C_{3}=p(p1)(p2).
For n + 1 points, if A_{1} and A_{n} have differenr colors, then A_{1}, ..., A_{n} can be colored in C_{n} ways, while A_{n+1} can be colored in p  2 ways. If A_{1} and A_{n} have the same color, then A_{1}, ..., A_{n} can be colored in C_{n1} ways and A_{n+1} can be colored in p  1 ways. So C_{n+1} = (p2)C_{n} +(p1)C_{n1} with n > 2 (*)
(*) can be written as:
C_{n+1} + C_{n} = (p  1) (C_{n} + C_{n1})
=> C_{n+1} + C_{n} = (p  1)^{n2} (C_{3} + C_{2}) = p(p  1)^{n}.
By induction we infer that C_{n} = (p  1)^{n} + (1)^{n} (p  1)

FA KAKALOTS 09/02/2018 at 21:43
Let A0 be the subset of S = {1, 2, ...3000} containing all numbers of the form 4nk, where n is a nonnegative integer and k is an odd positive integer. Then no two elements of A0 have ratio 2. A simple count shows A0 has 1999 elements. Now for each x∈ A0, form a set Sx={x,3x}∩S. Note the union of all Sx's contains S,S0, by the Dirichlet principle, any subset of S having more than 1999 elements must contain a pair in some Sx, hence of ratio 2. So no subset of 2000 numbers in S has the property.

london 11/04/2017 at 08:50
Let A_{0} be the subset of S = {1, 2, ...3000} containing all numbers of the form 4^{n}k, where n is a nonnegative integer and k is an odd positive integer. Then no two elements of A_{0} have ratio 2. A simple count shows A_{0} has 1999 elements. Now for each \(x\in\) A_{0,} form a set \(S_x=\left\{x,3x\right\}\cap S\). Note the union of all \(S_x\)'s contains \(S,S_0\), by the Dirichlet principle, any subset of S having more than 1999 elements must contain a pair in some \(S_x\), hence of ratio 2. So no subset of 2000 numbers in S has the property.