Number of non negative integer solutions to this inequality

[Woolyabyss]

Homework Statement

How many non-negative integer solutions are there to the equation
x1 + x2 + x3 + x4 + x5 < 11,
(i)if there are no restrictions?
(ii)How many solutions are there if x1 > 3?
(iii)How many solutions are there if each xi < 3?

Homework Equations

N/A

The Attempt at a Solution

(i) inequality equivalent to equality x1 + x2 + x3 + x4 + x5 + x6 = 10
(n+k-1)choose(k-1) = (10+6-1)choose(6-1) = 3003

(ii) if x1 > 3 ------------> x2 + x3 + x4 + x5 < 8
equality equivalent x2 + x3 + x4 + x5 +x6 = 7
again (7+5-1)choose(4) = 330

(iii)
Its this part I'm not certain about

like in (ii) let x1 instead be > 3 we find 7+5-1)choose(4) = 330

now let x1 and x2 be > 3
we find x3 + x4 + x5 < 5
----> x3 + x4 + x5 +x6 = 4
(4+4-1)choose(4-1) =35

finally let x1,x2 and x3 >3
then x4 +x5 <2
equality x4 + x5 + x6 =1
(1+3-1)choose(2) = 3

answer: 3003 - 35 - 330 - 3 = 2635

any help would be appreciated.

 

[haruspex]

You made a small error in (ii). If x1>3 then x1>=?
(iii) is actually the easiest of the questions. If every term is under 3, and there are only 5 terms...

 

[Woolyabyss]

You made a small error in (ii). If x1>3 then x1>=?
(iii) is actually the easiest of the questions. If every term is under 3, and there are only 5 terms...

is this correct?
(ii)
If x1>3 then x1 >= 4
if x1 >= 4 ------------> x2 + x3 + x4 + x5 < 7
equality equivalent x2 + x3 + x4 + x5 + x6 = 6
(6+5-1)choose(5-1)
(iii) since xi < 3
each term can be 0,1 or 2
so 3^5

 

[haruspex]

is this correct?
(ii)
If x1>3 then x1 >= 4
if x1 >= 4 ------------> x2 + x3 + x4 + x5 < 7
equality equivalent x2 + x3 + x4 + x5 + x6 = 6
(6+5-1)choose(5-1)
(iii) since xi < 3
each term can be 0,1 or 2
so 3^5

Looks right to me.

 

[Ray Vickson]

Homework Statement

How many non-negative integer solutions are there to the equation
x1 + x2 + x3 + x4 + x5 < 11,
(i)if there are no restrictions?
(ii)How many solutions are there if x1 > 3?
(iii)How many solutions are there if each xi < 3?

Homework Equations

N/A

The Attempt at a Solution

(i) inequality equivalent to equality x1 + x2 + x3 + x4 + x5 + x6 = 10
(n+k-1)choose(k-1) = (10+6-1)choose(6-1) = 3003

(ii) if x1 > 3 ------------> x2 + x3 + x4 + x5 < 8
equality equivalent x2 + x3 + x4 + x5 +x6 = 7
again (7+5-1)choose(4) = 330

(iii)
Its this part I'm not certain about

like in (ii) let x1 instead be > 3 we find 7+5-1)choose(4) = 330

now let x1 and x2 be > 3
we find x3 + x4 + x5 < 5
----> x3 + x4 + x5 +x6 = 4
(4+4-1)choose(4-1) =35

finally let x1,x2 and x3 >3
then x4 +x5 <2
equality x4 + x5 + x6 =1
(1+3-1)choose(2) = 3

answer: 3003 - 35 - 330 - 3 = 2635

any help would be appreciated.

You have part (i) wrong: the totality of non-negative integer solutions to $x_1+x_2+x_3+x_4+x_5 < 11$ is all the solutions to $\sum x_i = 0$ plus all the solutions to $\sum x_i = 1$, plus ... plus all the solutions to $\sum x_i = 9$ plus all the solutions to $\sum x_i = 10$. These all have sums < 11, as requested.

The solutions to other parts will be similarly affected.

 

[haruspex]

You have part (i) wrong: the totality of non-negative integer solutions to $x_1+x_2+x_3+x_4+x_5 < 11$ is all the solutions to $\sum x_i = 0$ plus all the solutions to $\sum x_i = 1$, plus ... plus all the solutions to $\sum x_i = 9$ plus all the solutions to $\sum x_i = 10$.

Did the trick of introducing a sixth unknown to take up the slack not resolve that?

 

[Ray Vickson]

Did the trick of introducing a sixth unknown to take up the slack not resolve that?

Yes, it does, but I missed that!