How many ways can you find natural numbers which add up to 10 if the order matters. Count 10 by itself as 1 way.

Assuming that we cannot repeat any natural number....

10       ( 1 way)

9, 1     ( 2 ways)

8, 2     (2 ways)

7, 3     (2 ways)

6, 4     (2 ways)

7, 2 , 1   (6 ways)

6, 3 , 1   (6 ways)

5, 4, 1   (6 ways)

4, 3, 2 , 1    (24 ways)

So....   1 + 4(2)  + 3(6)  + 24    =     51 ways

OK, The partitions of(10) =42 distinct partitions as follows:

This is called the "Partition Function = P(10)"

10 = 10
9 + 1 = 10
8 + 2 = 10
8 + 1 + 1 = 10
7 + 3 = 10
7 + 2 + 1 = 10
7 + 1 + 1 + 1 = 10
6 + 4 = 10
6 + 3 + 1 = 10
6 + 2 + 2 = 10
6 + 2 + 1 + 1 = 10
6 + 1 + 1 + 1 + 1 = 10
5 + 5 = 10
5 + 4 + 1 = 10
5 + 3 + 2 = 10
5 + 3 + 1 + 1 = 10
5 + 2 + 2 + 1 = 10
5 + 2 + 1 + 1 + 1 = 10
5 + 1 + 1 + 1 + 1 + 1 = 10
4 + 4 + 2 = 10
4 + 4 + 1 + 1 = 10
4 + 3 + 3 = 10
4 + 3 + 2 + 1 = 10
4 + 3 + 1 + 1 + 1 = 10
4 + 2 + 2 + 2 = 10
4 + 2 + 2 + 1 + 1 = 10
4 + 2 + 1 + 1 + 1 + 1 = 10
4 + 1 + 1 + 1 + 1 + 1 + 1 = 10
3 + 3 + 3 + 1 = 10
3 + 3 + 2 + 2 = 10
3 + 3 + 2 + 1 + 1 = 10
3 + 3 + 1 + 1 + 1 + 1 = 10
3 + 2 + 2 + 2 + 1 = 10
3 + 2 + 2 + 1 + 1 + 1 = 10
3 + 2 + 1 + 1 + 1 + 1 + 1 = 10
3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
2 + 2 + 2 + 2 + 2 = 10
2 + 2 + 2 + 2 + 1 + 1 = 10
2 + 2 + 2 + 1 + 1 + 1 + 1 = 10
2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 10
2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10

Now you can re-arrange these partitions into any order you want such as:3+3+2+2, 3+2+2+3, 2+2+3+3, 2+3+2+3.......etc. I would estimate that you would have more than 100 such re-arrangements.

For each of the partitioned sets, calculate the unique permutations using N!/((d₁!)* (d₂!) *...(d_z!)) Where N is the number of elements and d_z is the number of repeated elements (d) of each kind (_Z).  Ex:  for 4,2,2,1,1  it’s 5!/((2!)*(2!)) = 30. This represents 30 uniquely identifiable arrangements.  

\(\begin{array}{|r|r|r|r|r|r|r|r|r|} \hline \overset {..}\smile &G & I & N & G & E & R&A &L &E & \overset {..}\smile \\ \hline 10 & & & & & & & & & & 1\\ \hline 9 & 1 & & & & & & & & & 2 \\ \hline 8 & 2 & & & & & & & & & 2 \\ \hline 8 & 1 & 1 & & & & & & & & 3 \\ \hline 7 & 3 & & & & & & & & & 2 \\ \hline 7 & 2 & 1 & & & & & & & & 6 \\ \hline 7 & 1 & 1 & 1 & & & & & & & 4 \\ \hline 6 & 4 & & & & & & & & & 2 \\ \hline 6 & 3 & 1 & & & & & & & & 6 \\ \hline 6 & 2 & 2 & & & & & & & & 3 \\ \hline 6 & 2 & 1 & 1 & & & & & & & 12 \\ \hline 6 & 1 & 1 & 1 & 1 & & & & & & 5 \\ \hline 5 & 5 & & & & & & & & & 1 \\ \hline 5 & 4 & 1 & & & & & & & & 6 \\ \hline 5 & 3 & 2 & & & & & & & & 6 \\ \hline 5 & 3 & 1 & 1 & & & & & & & 12 \\ \hline 5 & 2 & 2 & 1 & & & & & & & 12 \\ \hline 5 & 2 & 1 & 1 & 1 & & & & & & 20 \\ \hline 5 & 1 & 1 & 1 & 1 & 1 & & & & & 6 \\ \hline 4 & 4 & 2 & & & & & & & & 3 \\ \hline 4 & 4 & 1 & 1 & & & & & & & 6 \\ \hline 4 & 3 & 3 & & & & & & & & 3 \\ \hline 4 & 3 & 2 & 1 & & & & & & & 24 \\ \hline 4 & 3 & 1 & 1 & 1 & & & & & & 20 \\ \hline 4 & 2 & 2 & 2 & & & & & & & 4 \\ \hline 4 & 2 & 2 & 1 & 1 & & & & & & 30 \\ \hline 4 & 2 & 1 & 1 & 1 & 1 & & & & & 30 \\ \hline 4 & 1 & 1 & 1 & 1 & 1 & 1 & & & & 7 \\ \hline 3 & 3 & 3 & 1 & & & & & & & 4 \\ \hline 3 & 3 & 2 & 2 & & & & & & & 6 \\ \hline 3 & 3 & 2 & 1 & 1 & & & & & & 30 \\ \hline 3 & 3 & 1 & 1 & 1 & 1 & & & & &15 \\ \hline 3 & 2 & 2 & 2 & 1 & & & & & & 20 \\ \hline 3 & 2 & 2 & 1 & 1 & 1 & & & & & 60 \\ \hline 3 & 2 & 1 & 1 & 1 & 1 & 1 & & & & 42 \\ \hline 3 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & & 8 \\ \hline 2 & 2 & 2 & 2 & 2 & & & & & & 1 \\ \hline 2 & 2 & 2 & 2 & 1 & 1 & & & & & 15 \\ \hline 2 & 2 & 2 & 1 & 1 & 1 & 1 & & & & 35 \\ \hline 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & & & 28 \\ \hline 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & 9 \\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \overset {..}\smile &G & I & N & G & E & R&A &L &E & \downarrow \\ \hline \end{array}\\ \small \text {Sum of all unique permutations } \hspace{2.85cm} \fbox {512}\\\)

EDIT: Corrected Errors

GA

this is how i tried to solve this question:

the question is: what is the number of partitions of the number 10 where the order matters, where a partition of 10 is a set of some positive natural numbers that add up to 10. We can rewrite the question as:

find the sum: (number of sets of 1 positive number where the order matters and the numbers add up to 10)+(number of sets of 2 positive numbers where the order matters and the numbers add up to 10)+...+(number of sets of 10 positive numbers where the order matters and the numbers add up to 10).

we need to find the number of sets of n positive numbers where the order matters and the numbers add up to 10.

using the stars and bars method the answer to that question is 10-1 choose n-1=9 choose n-1, so the answer to the question "what is the number of partitions of the number 10 where the order matters" should be

(9 choose 0)+(9 choose 1)+(9 choose 2)+...+(9 choose 9)=2⁹=512

this is my answer, but i see that other users got different answers, so i probably missed something

can someone please tell me what's wrong with my answer?

thanks

There are a couple of mistakes in the above breakdown:

She has listed only 41 partitions instead of 42. The omitted partition is first one on the list of 42 partitions: 10=10, which counts as 1 permutation. Then there is a mistake in her calculation of:

4, 2, 2, 2 =4 permutations NOT 6 as she has it listed.

So, taking her total of 513 + 1 - 2 = 512 total number of permutations, which agrees with your calculation.

Comments on the Stars & Bars Method

The Stars and Bars method solves this directly.  It’s quite cool, because it’s not intuitively obvious that it would.   Why does the sum of a series of the relatively simple S&B, (which is just nCr with adjusted parameters), equal the solution of a bifurcated solution process, where the Partitions by decomposition of a number into all sets, and then summing the permutation count of each partitioned set? 

Though it’s not intuitively obvious, it becomes obvious (sort of) after dissecting the equations. This is like finding Wally. It may take a while to find him, but once you do then you will (usually) always see him. There are many equations and formulas like this, especially in combinatorics.

Math texts will sometimes introduce procedures requiring the use of two or more formulas and then give an already introduced formula as the direct solution. “We leave it to the student to solve the relation.”  Lancelot Link did this to me all the time.  .

Occasionally I could solve them, but usually not.  Lancelot would usually explain these relationships, but sometimes I would get this as a response: “You know I never figured that one out. I was hoping you would because I’m very curious. 

You might now see why it was the School of Mathematics and Trolling.

GA