spring vale
there might be a better way to do this but brute force is a reasonable approach
7111 (4 permutations)
6211 (12 permutations)
5311 (12 permutations)
5221 (12 permutations)
4411 (6 permutations)
4321 (24 permutations)
4222(4 permutations)
3322 (6 permutations)
queen drum
Apply Stars and bars
sick oak
just give me result i beg you
queen drum
just give me result i beg you
!nosols
idle isleBOT
just give me result i beg you
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
sick oak
I didnt study this in my country
Give me a more matematica approach if you want me to solve it
queen drum
I'll give you an example:
If you have three numbers greater than 0 and you want the sum to be 5.
you assume a row of 5 stars
* * * * *
If you put two separators, | |
you would have something like:
*| * *| * * -> 1-2-2
* * *| *| * -> 3-1-1
which is just what you are looking for, so how many separators do you need?
3 numbers -1 = 2
And in how many places can you put the dividers:
*_*_*_*_* -> 4
On the edges not because they have to be greater than 0
So the sum you are looking for -1
5-1 = 4 places
So the calculation boils down to: How many ways can we place 2 objects in 4 places?
4C2 = 6
spring vale
I get what you're saying but that doesn't match my brute force method. Which solutions did I miss?
queen drum
For example when the first number is 1 or 2.
Ah now I understand what you did.
Letme find one
3331
spring vale
ty
fresh oak
.solved