# Thread: How many ways can the number 50 be witten as the sum of 10 non-negative, even number?

1. ## How many ways can the number 50 be witten as the sum of 10 non-negative, even number?

Hi,

Please help me to find out: in how many ways the number 50 can be witten as the sum of 10 non-negative, even numbers?

Thanks alot.

2. Hello, gravity2910!

In how many ways the number 50 can be witten
as the sum of 10 nonnegative, even numbers?
I hope the order of the numbers is important. .**
For example: $\{1,1,1,1,1,1,1,1,1,41\}$ is different from $\{41,1,1,1,1,1,1,1,1,1\}$

Consider fifty 1's in a row.

Group them in pairs .... and we have twenty-five 2's in a row.

. . . . . $\underbrace{\boxed{2}\;\;\boxed{2}\;\;\boxed{2}\;\ ;\boxed{2}\;\;\hdots\;\;\boxed{2}}_{\text{t\!went\ !y-five pairs}}$

There are 24 spaces between these pairs.

We will insert 9 dividers into those spaces.

For example: . $\boxed{2}\;\boxed{2}\;\bigg|\;\boxed{2}\;\bigg| \; \boxed{2}\;\boxed{2}\; \boxed{2}\;\boxed{2}\;\bigg| \;\hdots \; \bigg|\;\boxed{2}\;\boxed{2}\;\boxed{2}$

. . . . . . . . . . represents the sum: . $4 + 2 + 8 + \hdots + 6$

Therefore, there are: . $_{24}C_9 \;=\;{24\choose9} \;=\;\frac{24!}{9!\,15!} \;=\;\boxed{1,\!307,\!504}$ ways.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If the order of the numbers is not important,
. . the problem is far more complex.

Trust me on this!

3. Thank you very much, Soroban. Your help is great. This answer is very clear and easy to understand. I really appreciate it.

4. Originally Posted by gravity2910
Please help me to find out: in how many ways the number 50 can be witten as the sum of 10 non-negative, even numbers?
Does that answer account for the solution 0+0+0+0+0+0+0+0+0+50=50?
Those are all nonnegative even numbers.

If order matters is this a solution 2+0+4+0+6+0+8+0+0+30=50?
If so how is it counted?

Are we to assume that instead of nonnegative we read positive even numbers?
Then 2+4+6+8+10+2+8+2+6+2=50 is a solution and would be counted.

The number 25 can be written in 1455 ways using 10 or fewer summands.
Assume that order makes no difference and use the term nonnegative.
Then we get, one sum, 25=25.
That would count the solution 0+0+0+0+0+0+0+0+0+50=50.
Noting that 1+2+3+4+15=25, this counts 2+0+4+0+6+0+8+0+0+30=50.
Allowing zero to count as an even and order makes no difference there are 1455 ways.

If we read ten positive even numbers then the answer is 164.
25 can be written using exactly 10 summands in 164 ways.

Mind you I don’t know exactly how to read this question.
It could even mean "ten different even numbers"?

5. The above answer doesn't count for the case like 0+0+....+50. I think soroban counted only with positive and even numbers. For non-negative and even numbers, the answer should be:

6. I have another question that is slightly different to this question but causes some problems: in How many ways can the number 51 be witten as the sum of 11 positive, odd numbers? Could anyone help me??

7. Originally Posted by gravity2910
The above answer doesn't count for the case like 0+0+....+50. I think soroban counted only with positive and even numbers. For non-negative and even numbers, the answer should be:
But that answer means that order makes a difference.
Why do you think that should be the case?

8. Let imagine 2 2| 2 | |2 2 2|.......2 be x1|x2|.........|x10
because xi can be 0 then we can have xi=ai+1 in order to the dividers make sense.
x1+x2+.....+x10=n
If (a1,a2,......ar) is a solution then (a1+1,a2+1,.........,ar+1) is a solution to that problem with n replaced by n+r (since a1+1+a2+1+...+ar+1=n+r).
Like Soroban has explained: we have total: (n+r-1|r-1). In this case: (25+10-1|10-1)=(34|9).

These 2 formular (n-1|r-1) for the positive integers and (n+r-1|r-1) for non-negative integers are also written in some books. Exage 13, book: " A First Course in Probability" 7. edition of Sheldon Ross

9. Originally Posted by gravity2910
Let imagine 2 2| 2 | |2 2 2|.......2 be x1|x2|.........|x10
because xi can be 0 then we can have xi=ai+1 in order to the dividers make sense.
x1+x2+.....+x10=n
If (a1,a2,......ar) is a solution then (a1+1,a2+1,.........,ar+1) is a solution to that problem with n replaced by n+r (since a1+1+a2+1+...+ar+1=n+r).
Like Soroban has explained: we have total: (n+r-1|r-1). In this case: (25+10-1|10-1)=(34|9).

These 2 formular (n-1|r-1) for the positive integers and (n+r-1|r-1) for non-negative integers are also written in some books. Exage 13, book: " A First Course in Probability" 7. edition of Sheldon Ross
I know very well how and why you got the answer $\binom{25+10-1}{10-1}$.
But that means that order makes a difference. Why assume that?
That is my question to you.

10. sorry plato, i missunderstood you. I asked my prof. again and he said we assume the order makes the difference, so that our homework is more convenient.
By the way, how can you compute: The number 25 can be written in 1455 ways using 10 or fewer summands??

11. Originally Posted by gravity2910
By the way, how can you compute: The number 25 can be written in 1455 ways using 10 or fewer summands??
$\Pi (n,k)$ stands for number of ways to partition n into k or fewer summands.
$\Pi (n,k) = \left\{ {\begin{array}{ll} 1, & {k = 1} \\ {\Pi (n,k - 1) + \Pi (n,k - 1)}, & {1 < k < n} \\ {1 + \Pi (n,n - 1)}, & {k = n} \\ \end{array} } \right.$
As you can see it is a recusive function.
So it must programmed.