[jacks]

Q: The sum of \( 3 \) positive integer is \( 20. \) Then the probability that they form a Triangle is

i am assuming \( a+b+c=20,a,b,c \in \mathbb{N} \) using wlog \( a\leq b \leq c \)

for triangle \( a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10 \) and \( a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7 \)

so range of \( 7\leq c\leq 10 \)

For \( c=7. \) we have \( a+b=13, \) Then ordered pairs \( (6,7) \)

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

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

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

ordered pairs to form a triangle is \( 13 \)(favorable ways)

How do i calculate Total ways. please see it Thanks

[Luis Fuentes]

Hola

Q: The sum of \( 3 \) positive integer is \( 20. \) Then the probability that they form a Triangle is

i am assuming \( a+b+c=20,a,b,c \in \mathbb{N} \) using wlog \( a\leq b \leq c \)

for triangle \( a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10 \) and \( a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7 \)

so range of \( 7\leq c\leq 10 \)

For \( c=7. \) we have \( a+b=13, \) Then ordered pairs \( (6,7) \)

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

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

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

ordered pairs to form a triangle is \( 13 \)(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 \( 5 \) ordered triples of type \( (a,a,b) \). There are \( 3 \) ways of rearrange it.

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

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

The total ways are the number of integer solutions of:

\( x+y+z=20 \) with \( x,y,z\geq 1 \)

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

\( a+b+c=17 \)

that is \( \displaystyle\binom{3+\color{red}17\color{black}-1}{3-1} \)

Best regards.

CORRECTED

[Richard R Richard]

a	b	c	\( \triangle \)vs\( \cancel{\triangle} \)	order
1	1	18	\( \cancel{\triangle} \)	3
2	2	16	\( \cancel{\triangle} \)	3
3	3	14	\( \cancel{\triangle} \)	3
4	4	12	\( \cancel{\triangle} \)	3
5	5	10	\( \triangle \)	3
6	6	8	\( \triangle \)	3
7	7	6	\( \triangle \)	3
8	8	4	\( \triangle \)	3
9	9	2	\( \triangle \)	3
1	2	17	\( \cancel{\triangle} \)	6
1	3	16	\( \cancel{\triangle} \)	6
1	4	15	\( \cancel{\triangle} \)	6
1	5	14	\( \cancel{\triangle} \)	6
1	6	13	\( \cancel{\triangle} \)	6
1	7	12	\( \cancel{\triangle} \)	6
1	8	11	\( \cancel{\triangle} \)	6
1	9	10	\( \triangle \)	6
2	3	15	\( \cancel{\triangle} \)	6
2	4	14	\( \cancel{\triangle} \)	6
2	5	13	\( \cancel{\triangle} \)	6
2	6	12	\( \cancel{\triangle} \)	6
2	7	11	\( \cancel{\triangle} \)	6
2	8	10	\( \triangle \)	6
3	4	13	\( \cancel{\triangle} \)	6
3	5	12	\( \cancel{\triangle} \)	6
3	6	11	\( \cancel{\triangle} \)	6
3	7	10	\( \triangle \)	6
3	8	9	\( \triangle \)	6
4	5	11	\( \cancel{\triangle} \)	6
4	6	10	\( \triangle \)	6
4	7	9	\( \triangle \)	6
5	6	9	\( \triangle \)	6
5	7	8	\( \triangle \)	6

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

\( P_{\triangle} =\dfrac{13}{33} \)


with order and rotation

\( P_{\triangle} =\dfrac{3\cdot 5+6\cdot 8}{3\cdot 9+6\cdot 24}=\dfrac{63}{171}=\dfrac{7}{19} \)

[jacks]

Thanks Admin and Richard.

[Luis Fuentes]

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:

\( x+y+z=20 \) with \( x,y,z\geq 1 \)

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

\( a+b+c=17 \)

that is \( \displaystyle\binom{3+\color{red}17\color{black}-1}{3-1} \)

Best regards.

CORRECTED

Best regards.