This problem is about the sum of compositions of a ranging number n. Take n=3 and k=3. The composition of n into 3 parts (k=3) is (1,1,1). Then take n=n+1=4. Keep the k=3. The compositions of 4 into 3 parts are (1,1,2), (1,2,1), (2,1,1). Next increase the n again: n=n+1=5. The compositions of 5 into 3 parts are: (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1), (3,1,1). What is a formula for a sum of compositions, for a given, constant k, and n varying from k to k+m, m>0? This looks doable. The number of compositions of n into exactly k parts (k-composition) is given by the binomial coefficient $\binom {n-1}{k-1}$. So the question is about the sum of these binomial coefficients. If k=3 and k+m=6, then the sum is $\binom {2}{2} + \binom {3}{2} + \binom{4}{2} + \binom {5}{2} = 20$.
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Hint: $20 = { 6 \choose 3 }$ . Apply the hockey stick identity to your sum of binomial coefficients. – Calvin Lin Nov 29 '23 at 05:36
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I think the answer is $\binom {6} {3}$. I will try to prove it. – Stanley Taylor Nov 29 '23 at 13:17
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I googled the term 'hockey stick identity' and this answers my question. Thank you for help. – Stanley Taylor Nov 29 '23 at 13:28