# Number of non negative integer solutions to this inequality

• Woolyabyss
So the answer to (i) is## 1 + 5 + \binom{6}{2} + \binom{7}{3} + \binom{8}{4} + \binom{9}{5} + \binom{10}{6} ##In summary, there are 462 non-negative integer solutions to the equation x1 + x2 + x3 + x4 + x5 < 11, if there are no restrictions. If x1 > 3, there are 330 solutions. If each xi < 3, there are 2,635 solutions.
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?

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 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
haruspex said:
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

Woolyabyss said:
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.

Woolyabyss
Woolyabyss said:

## 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?

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.

Ray Vickson said:
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?

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

Yes, it does, but I missed that!

## 1. How do I calculate the number of non-negative integer solutions to an inequality?

The number of non-negative integer solutions to an inequality can be calculated by graphing the inequality and counting the number of integer points on or below the graph. Alternatively, you can set up a system of equations using variables to represent the non-negative integers and solve for the number of solutions.

## 2. What is the difference between non-negative and positive integers?

Non-negative integers include all whole numbers from 0 to infinity, while positive integers only include whole numbers greater than 0. Therefore, non-negative integers also include 0 as a possible solution, while positive integers do not.

## 3. How do I know if my solution is a non-negative integer?

If you are solving an inequality for non-negative integer solutions, your solution should be a whole number that is equal to or greater than 0. You can check this by substituting your solution into the original inequality and ensuring that it satisfies the inequality.

## 4. Can an inequality have an infinite number of non-negative integer solutions?

Yes, an inequality can have an infinite number of non-negative integer solutions. This can occur when the inequality has a slope of 0 or is parallel to the x-axis, meaning that any whole number on or below the graph would be a valid solution.

## 5. What if my inequality has more than one variable? How do I find the number of non-negative integer solutions?

If your inequality has multiple variables, you can use the same methods as mentioned in the first question. You may need to use algebraic techniques such as substitution or elimination to solve for one variable in terms of the others, and then count the number of integer solutions for that variable. Alternatively, you can use a graphing calculator or software to graph the inequality and count the number of integer points on or below the graph.

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