Question

Number of ordered sets {x,y,z}
such that x,y,z>0
and x+y+z <=12
x,y,z are positive integers,
plz help...

Answer 1

too many!!!
can you tell us what courses are you taking and what have you tried to solve this question so that we can discuss it

Answer 2

It seems as if this problem is equivalent to finding the number of ways to partition 12 into 3 parts.

Answer 3

Added to the partitions of 11, 10, 9, ..., 3

Answer 4

And then each partition (using the formal definition of partition) would have 3!=6 ways to order the elements, so we would then have to multiply by 6.

Answer 5

I must leave for a little while now, but I shall return later today to attempt to finish this problem.

Answer 6

Since each of \(x, y, z\) must be at least one, we can also think of this as adding the partitions of \(0, 1, 2, 3, 4, 5, 6, 7, 8, 9\) into no more than three parts.

Answer 7

0 can be partitioned only one way, and the permutations of \(x, y, z\) are all the same, so that's 1 way.

Answer 8

As for the rest of \(1, 2, 3, 4, 5, 6, 7, 8, 9\), If we partition them into 1 partition, 1 way to do that, but we can permute x, y, z in 3 ways in this case since two are the same. So this leaves us with another \(9*1*3=27\) ways with a total of \(28\).

Answer 9

We can now discard \(1\) since it can't be partitioned more than 1 way.

Answer 10

Now we need to figure out which ones we'll have to multiply our number of partitions by 3 since \(\{x, y, z\}\) will have two identical elements, and which ones we'll multiply our partitions by 6 because \(\{x, y, z\}\) won't have repeats.

Answer 11

As it turns out, if we're partitioning an odd number, we'll have to multiply by 6 always since 2 doesn't divide an odd number.

If we're partitioning an even number, we'll multiply one partition by 3 (same amount in both partitions), and the others by 6.

Answer 12

If we find the number of partitions for \(2, 3, 4, 5, 6, 7, 8, 9\) into 2 parts, we get \(1, 1, 2, 2, 3, 3, 4, 4\) respectively. That means we have \[3*(1+1+1+1) + 6*(1+1+2+2+3+3+4)=108\]more ways to make our ordered set \(\{x, y, z\}\). (or \(108+28=136\) total ways so far)

PS. The \(1+1+1+1\) comes from 4 even numbers we're partitioning, and \(1+1+2+2+3+3+4\) is the sum of the other partitions.

Answer 13

Now we finally need to find how many ways to partition \(3, 4, 5, 6, 7, 8, 9\) into 3 parts, and figure out which ones we multiply by 3, and which ones we multiply by 6.

Answer 14

Respectively it partitions into \(1, 1, 2, 3, 4, 5, 7\) parts. However, we have a new consideration now that we're partitioning into 3 parts. If we put the same amount into all three parts, we need to multiply by 1 instead of 3 or 6 because \(\{a,a,a\}\) is the same order no matter how we switch up the a′s. That means for 3, 6, 9 one of the partitions is multiplied by 1.

Answer 15

For this last step, we find that we need to calculate \[1*(1+1+1) + 3*(1+2+1+3+3+3)+6*(1+1+2+3)=84\]Which leaves us with a total of \[136+84=220\]Ordered sets \(\{x,y,z\}\) such that \(x,y,z>0\) and \(x+y+z \leq12\), \(x, y, z \in \mathbb{Z}\)

Answer 16

I'm not sure if there's a faster/better way to do this or not. That last step I did was by writing out some of the possible partitions myself.

Answer 17

10, 1,1 3

9,1,2 6

8,2,2 3
8,1,3 6

7,4,1 6
7,3,2 6

6,5,1 6
6,3,3 3
6,4,2 3
total =
42

Answer 18

the nos on the right are the noof ways of arranging the 3 nos on the left

Answer 19

i hope i have added it correctly and didnt miss any posss

Answer 20

You are missing a few. Namely,

5, 5, 2 3
5, 4, 3 6
4, 4, 4 1

Leaving you with a total of 52. However, you also need to count the ways to partition 3, 4, 5, 6, 7, 8, 9, 10, 11 as well. This will leave you with a much larger total.

Answer 21

I'd be willing to bet that once you've added all those in, you'll be left with a final total of 220.

Answer 22

lazy me
i missed a few!!