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« : 08/11/2019, 03:36:48 am »

Q: The sum of [texx]3[/texx] positive integer is [texx]20.[/texx] Then the probability that they form a Triangle is

i am assuming [texx]a+b+c=20,a,b,c \in \mathbb{N}[/texx] using wlog [texx]a\leq b \leq c[/texx]

for triangle [texx]a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10[/texx] and [texx]a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7[/texx]

so range of [texx]7\leq c\leq 10[/texx]

For [texx]c=7.[/texx] we have [texx]a+b=13,[/texx] Then ordered pairs [texx](6,7)[/texx]

For [texx]c=8.[/texx] we have [texx]a+b=12,[/texx] Then ordered pairs [texx](6,6),(5,7),(4,8)[/texx]

For [texx]c=9.[/texx] we have [texx]a+b=11,[/texx] Then ordered pairs [texx](5,6),(4,7),(3,8),(2,9)[/texx]

For [texx]c=10.[/texx] we have [texx]a+b=10,[/texx] Then ordered pairs [texx](5,5),(4,6),(3,7),(2,8),(1,9)[/texx]

ordered pairs to form a triangle is [texx]13[/texx](favorable ways)

How do i calculate Total ways. please see it Thanks
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« Respuesta #1 : 08/11/2019, 07:27:24 am »

Hola

Q: The sum of [texx]3[/texx] positive integer is [texx]20.[/texx] Then the probability that they form a Triangle is

i am assuming [texx]a+b+c=20,a,b,c \in \mathbb{N}[/texx] using wlog [texx]a\leq b \leq c[/texx]

for triangle [texx]a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10[/texx] and [texx]a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7[/texx]

so range of [texx]7\leq c\leq 10[/texx]

For [texx]c=7.[/texx] we have [texx]a+b=13,[/texx] Then ordered pairs [texx](6,7)[/texx]

For [texx]c=8.[/texx] we have [texx]a+b=12,[/texx] Then ordered pairs [texx](6,6),(5,7),(4,8)[/texx]

For [texx]c=9.[/texx] we have [texx]a+b=11,[/texx] Then ordered pairs [texx](5,6),(4,7),(3,8),(2,9)[/texx]

For [texx]c=10.[/texx] we have [texx]a+b=10,[/texx] Then ordered pairs [texx](5,5),(4,6),(3,7),(2,8),(1,9)[/texx]

ordered pairs to form a triangle is [texx]13[/texx](favorable ways)

How do i calculate Total ways. please see it Thanks

You can compute the number of all (unordered) triplets which form a triangle.

You have [texx]5[/texx] ordered triples of type [texx](a,a,b)[/texx]. There are [texx]3[/texx] ways of rearrange it.

And you have [texx]8[/texx] ordered triples of type [texx](a,b,c)[/texx]. There are [texx]6[/texx] ways of rearrange it.

So there are [texx]5\cdot 3+8\cdot 6=63[/texx] triplets (a,b,c) providing a triangle.

The total ways are the number of integer solutions of:

[texx]x+y+z=20[/texx] with [texx]x,y,z\geq 1[/texx]

This is equivalent to the number of non negative integer solutions of:

[texx]a+b+c=17[/texx]


that is [texx]\displaystyle\binom{3+\color{red}17\color{black}-1}{3-1}[/texx]

Best regards.

CORRECTED
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« Respuesta #2 : 08/11/2019, 08:08:20 am »

abc[texx]\triangle[/texx]vs[texx]\cancel{\triangle}[/texx]order
1118[texx]\cancel{\triangle}[/texx]3
2216[texx]\cancel{\triangle}[/texx]3
3314[texx]\cancel{\triangle}[/texx]3
4412[texx]\cancel{\triangle}[/texx]3
5510[texx]\triangle[/texx]3
668[texx]\triangle[/texx]3
776[texx]\triangle[/texx]3
884[texx]\triangle[/texx]3
992[texx]\triangle[/texx]3
1217[texx]\cancel{\triangle}[/texx]6
1316[texx]\cancel{\triangle}[/texx]6
1415[texx]\cancel{\triangle}[/texx]6
1514[texx]\cancel{\triangle}[/texx]6
1613[texx]\cancel{\triangle}[/texx]6
1712[texx]\cancel{\triangle}[/texx]6
1811[texx]\cancel{\triangle}[/texx]6
1910[texx]\triangle[/texx]6
2315[texx]\cancel{\triangle}[/texx]6
2414[texx]\cancel{\triangle}[/texx]6
2513[texx]\cancel{\triangle}[/texx]6
2612[texx]\cancel{\triangle}[/texx]6
2711[texx]\cancel{\triangle}[/texx]6
2810[texx]\triangle[/texx]6
3413[texx]\cancel{\triangle}[/texx]6
3512[texx]\cancel{\triangle}[/texx]6
3611[texx]\cancel{\triangle}[/texx]6
3710[texx]\triangle[/texx]6
389[texx]\triangle[/texx]6
4511[texx]\cancel{\triangle}[/texx]6
4610[texx]\triangle[/texx]6
479[texx]\triangle[/texx]6
569[texx]\triangle[/texx]6
578[texx]\triangle[/texx]6

There is not more ...I think so..
without ordering

[texx]P_{\triangle} =\dfrac{13}{33}[/texx]


with order and rotation

[texx]P_{\triangle} =\dfrac{3\cdot 5+6\cdot 8}{3\cdot 9+6\cdot 24}=\dfrac{63}{171}=\dfrac{7}{19}[/texx]
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« Respuesta #3 : Ayer a las 11:22:02 am »

Thanks Admin and Richard.
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« Respuesta #4 : Hoy a las 03:55:21 am »

Hi

Thanks Admin and Richard.

Note that I had a mistake. It has been corrected.

The total ways are the number of integer solutions of:

[texx]x+y+z=20[/texx] with [texx]x,y,z\geq 1[/texx]

This is equivalent to the number of non negative integer solutions of:

[texx]a+b+c=17[/texx]


that is [texx]\displaystyle\binom{3+\color{red}17\color{black}-1}{3-1}[/texx]

Best regards.

CORRECTED

Best regards.
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