Autor Tema: Positive integer triplets

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02 Mayo, 2019, 11:03 am
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jacks

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Number of positive integer triplets \( (x,y,z) \) in \( x+y+z=n, \) where \( x<y<z \)

02 Mayo, 2019, 04:21 pm
Respuesta #1

jbgg

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We consider the following sets

\(  A =\{ (x,y,z) : 0<x<y<z\ \text{ and }\ x+y+z=n\} \)

and

\(  B =\{ (x,y,z) : 0<x,y,z\ \text{ and }\ x\neq y\neq z\neq x\ \text{ and }\ x+y+z=n\}. \)

There is a relation between cardinals of both sets, which is

\( |B| = 3!\;|A|. \)

So we reduce the problem to count the number of element of set \( B \). In order to calculate the cardinal of \( B \), we decompose the set as

\( B = C \setminus (C_{12}\cup C_{13}\cup C_{23}\cup C_{123}), \)

where

\(  C = \{ (x,y,z) : 0<x,y,z\ \text{ and }\ x+y+z=n\}, \)

\(  C_{12} = \{ (x,y,z) : 0<x,y,z\ \text{ and }\ x=y\neq z\ \text{ and }\ x+y+z=n\}, \)

\(  C_{13} = \{ (x,y,z) : 0<x,y,z\ \text{ and }\ x=z\neq y\ \text{ and }\ x+y+z=n\}, \)

\(  C_{23} = \{ (x,y,z) : 0<x,y,z\ \text{ and }\ y=z\neq x\ \text{ and }\ x+y+z=n\}, \)

and

\(  C_{123} = \{ (x,y,z) : 0<x=y=z\ \text{ and }\ x+y+z=n\}. \)

We note that \(  C_{12},C_{13},C_{23},C_{123}\subset C \) and the sets \( C_{12},C_{13},C_{23},C_{123} \) are pairwise disjoint sets.

It turns out that \( |B| = |C| - |C_{12}| - |C_{13}| - |C_{23}| - |C_{123}| \).

We have to count the cardinal of \( B, C_{12},C_{13},C_{23} \) and \( C_{123} \).

It is easy (if not you can ask) to see that \( |C| = \binom{n-1}{2} \),

\(  |C_{12}| = |C_{13}| = |C_{23}| = \left\{\begin{array}{ll} k-1 & \text{ if } n\equiv 0 \pmod{2},\\ k & \text{ if } n\not\equiv 0 \pmod{2},\\\end{array}\right. \)

and


\(  |C_{123}| = \left\{\begin{array}{ll} 1 & \text{ if } n\equiv 0 \pmod{3},\\0 & \text{ if } n\not\equiv 0 \pmod{3},\\\end{array}\right. \)

where \(  n = 2k + r \) with \( r=0,1 \).

Please, check my arguments in case I made a mistake.

NOTE: \( |C_{12}| \) is incorrect, see the answer #4.

02 Mayo, 2019, 04:55 pm
Respuesta #2

jacks

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Thanks Jbgg. but how can i write ordered triplets \( (x,y,z) \) in \( n \) form.


02 Mayo, 2019, 05:43 pm
Respuesta #3

jbgg

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Thanks Jbgg. but how can i write ordered triplets \( (x,y,z) \) in \( n \) form.

Sorry, I don't get your question.

In order to solve the problem (number of triples) it must be calculated the number of elements of set \( A \) as defined in
We consider the following sets

\(  A =\{ (x,y,z) : 0<x<y<z\ \text{ and }\ x+y+z=n\} \)

[...]

03 Mayo, 2019, 02:10 am
Respuesta #4

jbgg

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Sorry, I made a mistake. I have just check it since I thought very fast.

Now I can confirm the solution is correct if the following is changed.

[...]

\(  |C_{12}| = |C_{13}| = |C_{23}| = \left\{\begin{array}{ll} n/2-2 & \text{ if } n\equiv 0 \pmod{2} \text{ and }n\equiv 0 \pmod{3},\\ n/2-1 & \text{ if } n\equiv 0 \pmod{2} \text{ and }n\not\equiv 0 \pmod{3},\\  (n-1)/2-1 & \text{ if } n\not\equiv 0 \pmod{2} \text{ and }n\equiv 0 \pmod{3},\\  (n-1)/2 & \text{ if } n\not\equiv 0 \pmod{2} \text{ and }n\not\equiv 0 \pmod{3},\\ \end{array}\right. \)

[...]

Sorry for the inconvenience.

05 Mayo, 2019, 05:20 am
Respuesta #5

jacks

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Thanks jbgg got it.