Question : Find the value of the following expression :

$$ \frac{\sum_{i=0}^{2024}\sum_{r=0}^{2024}(-1)^r{2024 \choose r}(2024-r)^i} {\sum_{r=0}^{2025}(-1)^r\binom{2025}{r}(2025-r)^{2025}} $$

I am not able to think of a way to manipulate and solve this algebraically , so I decided to test this on smaller summation boundaries to find a pattern in the answers For ex :

$$ \frac{\sum_{i=0}^{2}\sum_{r=0}^{2}(-1)^r{2 \choose r}(2-r)^i} {\sum_{r=0}^{3}(-1)^r\binom{3}{r}(3-r)^{3}} $$

On substituting 2024 with 2 and 2025 with 3 and solving the resultant expression (summing each case) I got the answer 2/6 which reduced to 1/3 . Here I observed that 3 is the upper boundary for the summation in the denominator , so I checked for 2 more cases which turned out to be quite tedious but :

  1. 2024 --> 3 and 2025 --> 4 Answer - 6/24 --> 1/4
  2. 2024 --> 4 and 2025 --> 5 Answer - 24/120 --> 1/5

A pattern is formed where the reciprocal of the upper boundary of the summation in denominator is the answer . There is also the pattern of factorials where 2 is factorial of 2 , 6 is of 3 , 24 is of 4 and 120 is of 5 . So I can guess the answer to my original question to be 1/2025 following the pattern , which also turns out to be the correct answer But this is definitely not how the question was intended to be solved , so I need a way to solve this with just algebraic manipulations in mind.

There could be a way to generalize this as :

$$ \frac{\sum_{i=0}^{k}\sum_{r=0}^{k}(-1)^r{k \choose r}(k-r)^i} {\sum_{r=0}^{k+1}(-1)^r\binom{k+1}{r}(k+1-r)^{k+1}} = \frac{k!}{(k+1)!}=\frac{1}{k+1} $$

  • $\begingroup$ Work on the numerator and denominator separately. Start with the denominator, try it for $ 2, 3, 4, 5$, then make a guess at the general formula and prove it. $\endgroup$
    – Calvin Lin
    May 24 at 15:22
  • $\begingroup$ How do you interpret $0^0$? (in the numerator, when $r=2024$ and $i=0$) $\endgroup$ May 24 at 15:31
  • $\begingroup$ Does this answer your question? Combinatorial proof of $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!$, using inclusion-exclusion $\endgroup$ May 24 at 15:53
  • $\begingroup$ @AnneBauval I have used 0^0 as 1 in my calculation $\endgroup$
    – satvik
    May 24 at 16:09
  • $\begingroup$ @AnneBauval I think the questions you referenced is quite similar and the answer n! is also what the numerator must result in my question , but in my question because of the double summation n is not a constant , rather it is the variable 'i' , and I think this does not directly work in my case . $\endgroup$
    – satvik
    May 24 at 16:13

1 Answer 1


Let $$N_k:=\frac{\sum_{i=0}^{k}\sum_{r=0}^{k}(-1)^r{k \choose r}(k-r)^i} {\sum_{r=0}^{k+1}(-1)^r\binom{k+1}{r}(k+1-r)^{k+1}} =\frac{\sum_{i=0}^ka_{i,k}}{a_{k+1,k+1}}$$ where $$a_{k,i}:=\sum_{r=0}^k(-1)^r\binom kr(k-r)^i.$$ In the proposed duplicate and posts linked to it, it is proved that $$a_{k,k}=k!.$$ In this other post and posts linked to it, it is proved that $$\forall i<k\quad a_{k,i}=0.$$ Therefore, $$N_k=\frac{k!+0}{(k+1)!}=\frac1{k+1}.$$


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