Let $n,k,\ell$ be integers for which $0\leq k<\ell \leq n-6$. For a fixed $n$, think of $k,\ell$ as being allowed to vary. I believe the values


$$(n-k-5)(k+1)(k+2)\binom n{k+3}~~~\text{and}~~~(n-\ell-5)(\ell+1)(\ell+2)\binom n{\ell+3}$$


are not equal. A proof they are not equal is the goal, but insight as to why they *might* not be equal would still be appreciated.


**Motivation**: The integer $(n-k-5)(k+1)(k+2)\binom n{k+3}$, when divided by $2(n-2)$, is the degree of the complex irreducible character of the symmetric group $S\_n$ corresponding to the partition $(n-k-3,3,1^k)$. I am a group theorist working on a general conjecture on character degrees; these character degrees *if they are distinct* will prove my conjecture to be true in the case of both the symmetric and the alternating groups.


**My work effort**:


$\bullet$ I ran computer tests for $6\leq n\leq 7000$. For those $n$, the integers are distinct (modulo my ability to write code, anyway).


$\bullet$ I thought maybe the set of prime divisors could be used to differentiate the values for distinct $k,\ell$, but there are a large number of values for $n$ where this thought failed.



Conjectured combinatorial non-equality
It looks like we may simply say which of them is greater.


Denoting $k+3=t$ and $f(t)=(n-k-5)(k+1)(k+2)\binom n{k+3}=(n-t-2)(t-1)(t-2)\binom n{t}$ we get $$\frac{f(t+1)}{f(t)}=
\frac{t (n - t - 3) (n - t)}{(t - 2) (t + 1) (n - t - 2)}=\frac{n-t-1-\frac{2}{n-t-2}}{t-1-\frac2{t}}.
$$
If $t<n/2$, this is greater than 1, if $t\geqslant n/2$, this is less than 1. Thus the function $f(t)$ increases up to $n/2$ and decreases after $n/2$. Therefore if $f(t)=f(s)$, $t<s$, we must have $t<n/2<s$. Let us compare $s$ and $n-t$, for this goal simplify the ratios
$$
\frac{f(n-t)}{f(t)}=\frac{n-t-1}{t-1}>1
$$
$$
\frac{f(n-t+1)}{f(t)}=1-
\frac{2 (n - 2) (n - 2 t + 1)}{(t - 2) (t - 1) (n - t - 2) (n - t + 1)}<1.
$$
Thus by monotonicity we should have $n-t<s<n-t+1$, a contradiction.


