Discrete Maths Problem?

Question

How many solutions are there to the equation,

where x1, x2, x3 are integers such that

x1 >= 1, x2 >= 2, x3 >= 1,

Any help and step by step procedure welcome, if you don't know don't answe. Thank you.

The equation is

x1 + x2 + x2 = 18

Answer 1

If x1 = 1, then there are 15 possible solutions.
If x1 = 2, then there are 14 possible solutions.
If x1 = 3, then there are 13 possible solutions.
...
If x1 = 15, then there is 1 possible solution.

The number of solutions is thus 15+14+13+...+1
That's 15 numbers, with the average number being 8, so the number of possible solutions is 15*8 = 120.

Answer 2

Write a C program with 3 loops for x1, x2 and x3.

I am too lazy to do it on my own:

for(x1=1; x<=15; x1++)
{
for(x1=2; x<=16; x1++)
{
for(x1=1; x<=15; x1++)
{
if( x1 + x2 + x3 == 18 )
counter++;
}
}
}

cout<

Answer 3

There are an infinite number of solutions as x1, x2 and x3 are all independent. Your "equation" defines an area in 3 dimensional space that is bounded on one side and unbounded on the other.

Answer 4

Yay, the assumption of inclusion-exclusion that a great style of people have not got an awareness of whilst counting. solid, A U B U C = a,b,c,d,e,g,h,ok,m,n, which has 10 factors, so in addition A n B = a,b,e, which has 3 factors, so 3 when you do the completed intersections, in simple terms be counted the quantity of things in each and every and exchange those values into the equation. It would desire to be a genuine equation in spite of everything the numbers are put in there.

Answer 5

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You may want to visit that page, lots of info there

Answer 6

Are you weak in English too?