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Find $\frac{{n\choose 1}(n-1)^3+{n\choose 3}(n-3)^3+...}{n^2(n+3)2^n}$

My attempt at this question was to first write the numerator as $\sum_{k=1}^{n} {n\choose 2k-1}(n-(2k-1))^3$ and then expanding this we would get expression like $\sum_{k=1}^{n} {n\choose 2k-1}(2k-1)^3$ which if I write as $\frac{1}{2}\sum_{r=0}^{n}{n\choose r}r^3$ I might be able to solve this using the repeated differentiation of $(1+x)^n$ , But overall I am not sure if my approach is correct the sum seemed to get more complicated , is there any more elegant ways to solve this ?

SOUMILI NAG
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2 Answers2

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You have $n$ people, of these you want to select an odd number of people to be in a committee, then from the non committee you choose a technical, marketing, and sales manager where one person can have more than one title. The number of possibilities are:

$$ \sum_{i\in\mathbb{N}}{\binom{n}{2i-1}\left(n-(2i-1)\right)^{3}} $$

Which is the numerator of your equation. Now let’s count in a different way. We choose the managers first then select an odd number of people from the non managers. The following are the sum of possibilities from the case when all managers are different, one person hold two titles, and one person hold three titles:

$$ 3!\cdot\binom{n}{3}\cdot 2^{n-4}+3\cdot2\cdot\binom{n}{2}\cdot 2^{n-3}+\binom{n}{1}\cdot 2^{n-2} $$

Here I use the well known fact that there are $2^{m-1}$ odd - subset of a set with $m$ members. The two expressions are equal since both count the same possibilities, only with different method

acat3
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  • @Rehza Adrian Tanuhajra this method is Awesome...but is there any way to do it using algebra? – SOUMILI NAG Aug 21 '21 at 06:08
  • @SOUMILINAG I think algebraically your idea of differentiation of $(1+x)^{n}$ is a good idea – acat3 Aug 21 '21 at 06:13
  • @SOUMILINAG try differentiating, then multiply by $x$, then repeat 2 more times. Then substitute $x=1$ & $x=-1$, the difference is twice your numerator – acat3 Aug 21 '21 at 06:18
  • The correct way. Btw it should be possibilities not probabilities – tryst with freedom Aug 21 '21 at 06:35
  • @Buraian the reason I prefer combinatorial proof haha good answer btw – acat3 Aug 21 '21 at 06:37
  • In the $\binom{n}{2}$ you should be multiplying by two instead of 3! right? also how did you come up with this – tryst with freedom Aug 21 '21 at 06:45
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    @Buraian the 2 is because we have two people so we want to choose who has two titles, the 3 is to choose whether the lone title is technical, marketing, or sales manager – acat3 Aug 21 '21 at 07:09
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Algebraic brute force way:

$$ S = \sum_{k=1}^n \binom{n}{2k+1} \left[ n- (2k+1) \right]^3= \sum_k \binom{n}{2k+1} \left[ \sum_j \binom{3}{j} (2k+1)^j n^{3-j}\right]= \sum_j \binom{3}{j} n^{3-j} \sum_k \binom{n}{2k+1} (2k+1)^j \tag{1}$$

We need to evaluate: $$\sum_k \binom{n}{2k+1} (2k+1)^j$$

Notice that $$x \frac{d}{dx} (1+x)^n= x\frac{d}{dx} \sum_k \binom{n}{k} x^k = \sum_k \binom{n}{k}k x^k$$

Call $\phi= x \frac{d}{dx}$ then:

$$ \phi^j \left[ (1+x)^n\right] = \sum_k \binom{n}{k} k^j x^k $$

Now, we just need to take odd part of both sides and evaluate at one, which is given as:

$$ \phi^j \left[ (1+x)^n - (1-x)^n \right]|_{x=1} = \sum_k \binom{n}{2k+1} (2k+1)^j $$

Hence (1) becomes:

$$ S= \sum_j \binom{3}{j} n^{3-j} \phi^j \left[ (1+x)^n - (1-x)^n \right]_{x=1}$$

It maybe noted that $\phi^j$ can be expanded into a linear combination of ordinary derivatives weighted by Stirling numbers. Refer

tryst with freedom
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