Find the number of integer solutions of x1 + x2 + x3 = 15

[mathproblems]

Find the number of integer solutions of x1 + x2 + x3 = 15 subject to the conditions given.
0 < = x1 < 6,
1 < = x2 < 9,
x3 >= 0

c(15+3-1,15) –[ C(6+3-1,6) + C (7+3-1,7)+C(14+3,14,3)-C(1+3-1,1)]

Is this correct?

Thank you.

 

[pka]

Find the number of integer solutions of x1 + x2 + x3 = 15 subject to the conditions given.
0 < = x1 < 6,
1 < = x2 < 9,
x3 >= 0
c(15+3-1,15) –[ C(6+3-1,6) + C (7+3-1,7)+C(14+3,14,3)-C(1+3-1,1)]

Frankly, I do not follow your reasoning there.
How many ways are there for \(\displaystyle x_1+x_2+x_3=15\text{ with }x_2\ge 1~?\)

Subtract from that number the number of cases where \(\displaystyle x_1\ge 6\); subtract the number of cases where \(\displaystyle x_2\ge 9\) then add back the number of cases in which \(\displaystyle x_1\ge 6\text{ and }x_2\ge 9\).

 

[mathproblems]

why x2>=1, what about x1>=0?

for x1: ---> since 0 < = x1 < 6, gives 6 possible integers for x1
then 15-6 =9
C(9+3-1,9)

for x2: ----> since 1 < = x2 < 9, gives 8 possible integers for x2
C(6+3,1,6)


for x3: ----> since x3 >= 0
C(14+3,14,3)

Please correct me. thank you.

 

[pka]

why x2>=1, what about x1>=0?
for x1: ---> since 0 < = x1 < 6, gives 6 possible integers for x1
then 15-6 =9
C(9+3-1,9)
for x2: ----> since 1 < = x2 < 9, gives 8 possible integers for x2
C(6+3,1,6)
for x3: ----> since x3 >= 0
C(14+3,14,3) Please correct me. thank you.

Please do not take this as a put down.
But I don't think you have any idea how this works.
How many ways are there for \(\displaystyle x_1+x_2+x_3=15\text{ where }x_2\ge 1\)?
The answer is \(\displaystyle \binom{14+3-1}{2}\). That is we put a 1 into \(\displaystyle x_2\) and then put fourteen 1's into the other three variables.

Do you get that?

 

[soroban]

Hello, mathproblems!

I found the answer by "counting".

Find the number of integer solutions of: .\(\displaystyle x + y + z \:=\: 15\)

subject to the conditions: .\(\displaystyle \begin{Bmatrix}0 \le x < 6 \\ 1 \le y < 9 \\ z \ge 0 \end{Bmatrix}\)

Suppose \(\displaystyle y = 8.\)
. . Then we have: .\(\displaystyle \boxed{\begin{array}{ccc} x & y & z \\ \hline 0 & 8 & 7 \\ 1 & 8 & 6 \\ 2 & 8 & 5 \\ 3 & 8 & 4 \\ 4 & 8 & 3 \\ 5 & 8 & 2 \end{array}} \;\;\hdots\;\text{ 6 solutions}\)


Similarly, there are 6 solutions each for \(\displaystyle y \:=\:7, 6, 5, 4, 3, 2, 1.\)

Therefore, there are: .\(\displaystyle 8 \times 6 \:=\:48\) solutions.

 

[mathproblems]

I have a simular problem in the book with the solution. Discrete Mathematics, 7ed by Richard Johnsonbaugh

So I was trying to solve mine the same way.

Find the number of solutions in integers to x1 + x2 + x3 + x4 = 12 satisfying 0 <= x1 <= 4, 0 <= x2< = 5, 0 <= x3 <= 8, and 0 <= x4 <= 9.
Answer in the book: C( 12 + 4 - 1, 12) - [ C( 7 + 4 - 1, 7) + C( 6 + 4 - 1, 6) + C( 3 + 4 - 1, 3) + C( 2 + 4 - 1, 2) - C( 1 + 4 - 1, 1)]

So I was trying to solve mine the same way.
:???:

 

[mathproblems]

Hi, It sounds right, but...I dont see z=0 in your 6 ways.

and I dont get it, how did we just pick x2? why not x1 or x3?

Thank you!!!

Hello, mathproblems!

I found the answer by "counting".


Suppose \(\displaystyle y = 8.\)
. . Then we have: .\(\displaystyle \boxed{\begin{array}{ccc} x & y & z \\ \hline 0 & 8 & 7 \\ 1 & 8 & 6 \\ 2 & 8 & 5 \\ 3 & 8 & 4 \\ 4 & 8 & 3 \\ 5 & 8 & 2 \end{array}} \;\;\hdots\;\text{ 6 solutions}\)


Similarly, there are 6 solutions each for \(\displaystyle y \:=\:7, 6, 5, 4, 3, 2, 1.\)

Therefore, there are: .\(\displaystyle 8 \times 6 \:=\:48\) solutions.

 

[mathproblems]

thank you for your help. why 3-1?

Please do not take this as a put down.
But I don't think you have any idea how this works.
How many ways are there for \(\displaystyle x_1+x_2+x_3=15\text{ where }x_2\ge 1\)?
The answer is \(\displaystyle \binom{14+3-1}{2}\). That is we put a 1 into \(\displaystyle x_2\) and then put fourteen 1's into the other three variables.

Do you get that?

 

[soroban]

Hello, math problems!

It sounds right, but ... I dont see \(\displaystyle z=0\) in your 6 ways. . You won't!

Note that: .\(\displaystyle 0 \le x < 6\)

That is: .\(\displaystyle x \:=\:5,4,3,2,1,0\)


My first table is when \(\displaystyle y = 8.\)

. . \(\displaystyle \boxed{\begin{array}{ccc} x & y & z \\ \hline 5 & 8 & 2 \\ 4 & 8 & 3 \\ 3 & 8 & 4 \\ 2 & 8 & 5 \\ 1 & 8 & 6 \\ 0 & 8 & 7 \end{array}}\)



My last (eighth) table is when \(\displaystyle y = 1.\)

. . \(\displaystyle \boxed{\begin{array}{ccc}x & y & z \\ \hline 5 & 1 & 9 \\ 4 & 1 & 10 \\ 3 & 1 & 11 \\ 2 & 1 & 12 \\ 1 & 1 & 13 \\ 0 & 1 & 14 \end{array}}\)


You see, \(\displaystyle z\) never equals zero (or one).

 

[pka]

I have a simular problem in the book with the solution. Discrete Mathematics, 7ed by Richard Johnsonbaugh
So I was trying to solve mine the same way.

As I said before, think of \(\displaystyle x_2\) as having a 1 already in it.
Then that leaves fourteen 1's to distribute.
\(\displaystyle \binom{16}{2}-\binom{10}{2}-\binom{8}{2}+1=48\)

 

[mathproblems]

yes. thank you everyone. I got the 48.