Probability dots and lines problem.

[charmedbeauty]

Homework Statement

How many solutions are there to the equation

where x_n is non negative.

x1+x2+x3+x4+x5=30

where x1,x2,x3,x4,x5<8?

Homework Equations

The Attempt at a Solution

well I thought the only way to ensure this is when the first 7 'dots' are placed, since if all the 'dots' are placed in let's say x1 then by the 8th dot it could also be arranged in x1, hence not satisfying the condition.

ie C(11,4) ... but the sltn says C(9,4).

which I thought was strange; doesn't that sltn imply that x_n can ≥8 when the first 5 dots are arranged?

and the only condition on x_n is that they are non negative... so they can be zero.

 

[Bread18]

Each [itex]x_n[/itex] can have a maximum of 7 dots. So we can rewrite it as [itex]y_n = 7 - x_n[/itex]. Subbing this into the eqn you will get the answer to be C(9,4)

 

[charmedbeauty]

Each [itex]x_n[/itex] can have a maximum of 7 dots.

yeah I get that...

we can rewrite it as [itex]y_n = 7 - x_n[/itex]. Subbing this into the eqn you will get the answer to be C(9,4)

into what eqn?

x1+x2+x3+x4+x5=30?

I really don't see how... I thought C(9,4) was the number of ways we can arrange the 'lines'?

if so I really don't see how this solution is correct, since if I throw away the first 23 dots then there are 7 remaining.

now anyway I arrange these will satisfy the condition, where x1,x2,x3,x4,x5<8. would it not?

I know that C(9,4) does satisfy the condition, but so should C(11,4)?

 

[Bread18]

What do you mean by "throw away the first 23 dots"?

Think of it as having 7 dots in each [itex]x_n[/itex] and then place 5 'anti dots' into any of the [itex]x_n[/itex]. This is the same as having [itex]x_1 + x_2 + x_3 + x_4 + x_5 = 5[/itex]. (5 'anti dots' and 4 lines).

 

[charmedbeauty]

What do you mean by "throw away the first 23 dots"?

Think of it as having 7 dots in each [itex]x_n[/itex] and then place 5 'anti dots' into any of the [itex]x_n[/itex]. This is the same as having [itex]x_1 + x_2 + x_3 + x_4 + x_5 = 5[/itex]. (5 'anti dots' and 4 lines).

ohh I think I get it now. so 'antidots' are just the remaining dots after 7 dots have been distributed in the 5 subcategories x1...xn. Since there is only 1 way of arranging 7 dots in each category therefore there are 5 'antidots' to be distributed is C(9,4). Is this the right line of thinking?

Thanks for the help Bread.

 

[Bread18]

Yeah that's right.