Multiple variable equation help needed please :)

[cremebrulay]

hi, firstly i apologise if this has been posted in the wrong section, but I have no idea what this is called in mathematical terms.

The problem i have is this:

I have 11 variables a,b,c,d,e,f,g,h,i,j,k

Each variable has a value of 1 to 6

I need to solve as follows:

a+b+c+d+e+f+g+h+i+j+k=z

where z is fixed eg 20
and a-k can be any value of 1-6 eg

a=2
b=3
c=1
d=1
e=1
f=1
g=5
h=2
i=2
j=1
k=1

such that: 2+3+1+1+1+1+5+2+2+1+1=20

obviously there are i suspect an awful lot of variations of a-k that add up to z and my with my very limited knowledge of equations i have no idea how to even start to solve this other than manually which will take forever.

Any help will be very gratefully recieved, even if it just to tell me what branch of mathematics this falls into so i can google it! thankyou in advance

 

[Steven G]

Hi,
I would say that this is a probability problem. Think of this problem as follows using z=20.

You have 20 dashes (_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ) and you want to partition it into 11 regions. So you want to make 10 breaks like

_ _| _ _| _| _ _ |_ |_| _ _| _ _ _| _ _| _ _| _ _. This corresponds to 2+2+1+2+1+1+2+3+2+2+2 =20. So how many places can you put a vertical line in? The answer is 19 (after the 1st dash, after the 2nd dash, ..., after the 19th dash). And you want to choose 10 places to put the vertical lines. So the answer is 19C10..... In general the answer is (Z-1)C10
For z=20 there are going to be I bet about 100,000 ways to do this. There is no real method to write down all of them. Just go in a common sense order.
Jomo

 

[Steven G]

If it's not a probability problem (as Jomo stated), then it's a silly problem!

Ready? There are (if z=20) 89,232 solutions!
From 1 1 1 1 1 1 1 1 1 5 6 to 6 5 1 1 1 1 1 1 1 1 1

However, if you keep a to k in ascending order there are only 23.
From 1 1 1 1 1 1 1 1 1 5 6 to 1 1 2 2 2 2 2 2 2 2 2

From where does this problem originate?

Denis, If you have the time can you please explain how you get 23 if in ascending order. Thank you very much.
Jomo

 

[cremebrulay]

If it's not a probability problem (as Jomo stated), then it's a silly problem!

Ready? There are (if z=20) 89,232 solutions!
From 1 1 1 1 1 1 1 1 1 5 6 to 6 5 1 1 1 1 1 1 1 1 1

However, if you keep a to k in ascending order there are only 23.
From 1 1 1 1 1 1 1 1 1 5 6 to 1 1 2 2 2 2 2 2 2 2 2

From where does this problem originate?

thanks Jomo and Denis. the problem is from a battle card game i play online where i can make various different decks to fight an enemy deck. i have simplified this into its basic maths theory in order to try and work things out. What i am really interested in is the caluculations or formulae that allowed you to work out that this problem had 89,232 solutions because the value of z also has a range of values, (i chose 20 as an example) and if i knew the formulae or how to work out the number of solutions i will then be able to apply that to a further model to calculate the various decks etc.
many thanks

 

[Steven G]

Analysis, more or less; like:
min of 2 1's; leaving only 9 2's as possible
3 1's: only 7 2'a and 1 3 possible
4 1's: 2 cases: 5 2's and 2 3's ; 6 2's and 1 4
and so on...
Confirmed by looper program...

n = 6
t = 20
p = ?
Evident that a formula is required, something like p = something in terms of n and t;
can't find one, can't "invent" one...yet!

OK, thanks

 

[Steven G]

thanks Jomo and Denis. the problem is from a battle card game i play online where i can make various different decks to fight an enemy deck. i have simplified this into its basic maths theory in order to try and work things out. What i am really interested in is the caluculations or formulae that allowed you to work out that this problem had 89,232 solutions because the value of z also has a range of values, (i chose 20 as an example) and if i knew the formulae or how to work out the number of solutions i will then be able to apply that to a further model to calculate the various decks etc.
many thanks

In general nCr= [n!]/[r!(n-r)!]. In case you are not familiar with factorials (!) we have n! 1*2*3...*n. So 5! =1*2*3*4*5=120.
For the record, 19C10 = 19!/[10!*9!]=92,378. I only say this so that if you work it out yourself and do not get the result that Denis got you will not think you are wrong. This is a very very rare mistake on Denis' part.