no of non negative solutions of equation a+b+c+...n variables=N where a<=b<=c<=d<=....

Asked Jul 25 '18 at 19:46

Active Jul 27 '18 at 08:42

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I need to find the number of non negative solutions of equation $\underbrace{a+b+c+ \dotsb }_{n \text{ variables}}=N$ where $a\le b \le c \le d \le \dotsb $.

What I have tried is:

Suppose $S(n)$ gives us the required solution.

For $n=1$, only one solution $(a=N)$, $S(1)=1$

for $n=2$, $a+b=N$. $a$ can take either $0$ or non zero. if a takes 0 then total solutions = $S(1)=1$; otherwise, a can take values from 1 to N/2 so total floor(N/2) possible solutions. Hence $S(2)=S(1)+N/2 = 1 + N/2$

for $n=3, a+b+c=N$, a can take either 0 or non zero if a takes 0 then total solutions = $S(2)$ otherwise , a can take values from 1 to N/3 so total floor(N/3) possible solutions. Hence $S(3)= S(2) + N/3 = 1 + N/2 + N/3$

Similarly for $n=n$, $S(n)=1+ N/2 + N/3 + N/4 + ....... + N/n$

Can anyone tell whether the above formula is right or wrong?

edited Jul 27 '18 at 08:42

Peter Taylor

13,425

asked Jul 25 '18 at 19:46

Devendra Verma

3

Here's a MathJax tutorial :) – Shaun Jul 25 '18 at 19:48
1

Why downvoted. I was even editing the question – Devendra Verma Jul 25 '18 at 19:56
This seems to be related (that $\le n$ looks weird) to "The number of weak compositions of $N$ into exactly $n$ parts" – mvw Jul 25 '18 at 19:57
The question was unclear until just now. – Shaun Jul 25 '18 at 19:57
2

For $n=3,N=5$ you get $1+\lfloor \frac 52\rfloor +\lfloor \frac 53\rfloor=4$, no? But we have $(0,0,5),(0,1,4),(0,2,3),(1,1,3),(1,2,2)$. – lulu Jul 25 '18 at 20:02
@lulu yes it is wrong – Devendra Verma Jul 25 '18 at 20:04
@mvw It means that I dont need to worry about constraint. I need to find only possible solution (without any repetition) and count that? – Devendra Verma Jul 25 '18 at 20:07
Have you had a look here? – mvw Jul 25 '18 at 20:08
@mvw Oh man, I was thinking for very long and didn't realized that it is the same thing as distributing n identical balls into k identical boxes. Correct me if I am wrong. – Devendra Verma Jul 25 '18 at 20:15
Sounds reasonable, but I tend to make mistakes in combinatorics. – mvw Jul 25 '18 at 20:18
Here is an example https://math.stackexchange.com/questions/2623657/number-of-positive-integral-solutions-of-abcde-20-such-that-abcde-an/2623713#2623713 – rtybase Jul 26 '18 at 00:07
And another one https://math.stackexchange.com/questions/2862475/find-the-number-of-ways-to-choose-n-non-negative-numbers-that-sum-up-to-s-and – rtybase Jul 26 '18 at 00:08
The terminology you need to find an answer is "partitions of N into at most n parts", although the most standard variable naming would be "partitions of n into at most k parts". – Peter Taylor Jul 27 '18 at 08:48

no of non negative solutions of equation a+b+c+...n variables=N where a<=b<=c<=d<=....

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