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combinations

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Huỳnh Anh Phương
30/09/2018 at 11:13
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1
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Only the psycho will answer this question:

How many combinations Billy can make if Billy have 3 shirts, 8 trousers, 4 coats?

Who make the right answer and fastest I will T_I_C_K

Hurry up guys !!!

combinations

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    Quoc Tran Anh Le Coordinator 01/10/2018 at 07:35
    There are: 3*8*4=96 combinations that Billy can make. Huỳnh Anh Phương selected this answer.

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An Duong
10/04/2017 at 08:25
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2
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 Let A1; A2; ...; An be points on a circle. Find the number of possible colorings of these points with p colors, p > 1; such that any two neighboring points have distinct colors

combinations

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    FA KAKALOTS 09/02/2018 at 21:43

    Let Cn be the answer for n points. We have C1 = p, C2 = p(p-1) and C3=p(p-1)(p-2).

    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 Cn-1 ways and An+1 can be colored in p - 1 ways. So Cn+1 = (p-2)Cn +(p-1)Cn-1 with n > 2 (*)

    (*) can be written as:  

         Cn+1 + Cn = (p - 1) (Cn + Cn-1)

    => Cn+1 + Cn = (p - 1)n-2 (C3 + C2) = p(p - 1)n.

    By induction we infer that Cn = (p - 1)n + (-1)n (p - 1)

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    london 11/04/2017 at 08:43

    Let Cn be the answer for n points. We have C1 = p, C2 = p(p-1) and C3=p(p-1)(p-2).

    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 Cn-1 ways and An+1 can be colored in p - 1 ways. So Cn+1 = (p-2)Cn +(p-1)Cn-1 with n > 2 (*)

    (*) can be written as:  

         Cn+1 + Cn = (p - 1) (Cn + Cn-1)

    => Cn+1 + Cn = (p - 1)n-2 (C3 + C2) = p(p - 1)n.

    By induction we infer that Cn = (p - 1)n + (-1)n (p - 1)


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An Duong
10/04/2017 at 08:27
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2
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(1987 Austrian-Polish Math Competition)

Does the set {1;2;...;3000} contain a subset A consisting of 2000 numbers such that x \(\in\)A implies 2x \(\notin\) A? 

combinations

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    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.

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    london 11/04/2017 at 08:50

    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 \(x\in\) A0, 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.


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An Duong
10/04/2017 at 08:29
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(1989 Polish Math Olympiad)

Suppose a triangle can be placed inside a square of unit area in such a way that the center of the square is not inside the triangle. Show that one side of the triangle has length less than 1.

 

combinations


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An Duong
10/04/2017 at 08:30
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The cells of a 7 x 7 square are colored with two colors. Prove that there exist at least 21 rectangles with vertices of the same color and with sides parallel to the sides of the square.

 

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