How many ways to put 20 indistinguishable objects into 4 groups?

GalwayMathsStudent

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The question first says put 20 indistinguishable objects into 4 groups. How many ways can this be done?

This part was pretty easy ( I think) I just used the formula (n+r-1)c(r-1) and got the answer 1771.

The second part asked How many ways if not all objects need to be put in a group?
I am pretty sure this is just the sum from 1 to 20 indistinguishable objects into 4 groups.
After some manipulation I got it down to =(1/6) sum (from k=1 to 20){ ((3+k)!)/(k!)}
which I then simplified to = (1/6) sum (from k=1 to 20) {k^3+6k^2+11k+6}


I just can't see a way to get an answer fast. Sure I could just add 20 values from my calculator but I feel like I am missing some shortcut or easier approach to this problem.
Any Feedback would be appreciated
 
The question first says put 20 indistinguishable objects into 4 groups. How many ways can this be done? This part was pretty easy ( I think) I just used the formula (n+r-1)c(r-1) and got the answer 1771.

The second part asked How many ways if not all objects need to be put in a group?
The answer you gave for the first question assumes that the groups are distinguishable. That is not stated in the statement of the question. Does that need to be added? As stated, the answer would be the number of summands of twenty into four or fewer cells. Example: {10, 5, 3,2} or {10,5,5} or {20} etc.

Then for the next question, do that for each number one to twenty.
 
The answer you gave for the first question assumes that the groups are distinguishable. That is not stated in the statement of the question. Does that need to be added? As stated, the answer would be the number of summands of twenty into four or fewer cells. Example: {10, 5, 3,2} or {10,5,5} or {20} etc.

Then for the next question, do that for each number one to twenty.

Thanks for the reply pka. The groups in the question were distinguishable (I think). Sorry but I didn't really understand
this part of your post " As stated, the answer would be the number of summands of twenty into four or fewer cells. Example: {10, 5, 3,2} or {10,5,5} or {20} etc."


Also I know I could just sum up all the sum up all the possible values (from k=1-20 in my attempt) for the second question but that seems slow/awkward and I was looking to see if there was a shortcut using some rule/formula.

Although now that you mention it how would the groups being indistinguishable change the answer/formula?
 
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