Are these two combinatorially-defined sets of integers disjoint? Fix an integer $n\geq 8$.  For each integer $i\leq n/2$, denote by $X_i$ the set
        $$X_i = \left\{ \frac{n-i+1-k}{n-i+1}\binom nk\binom{k-1}{i-1} ~\middle|~~ i\leq k\leq n-i\right\}.$$
The question:
Are the sets $X_1$ and $X_2$ disjoint?
The reason:
The product $\frac{n-i+1-k}{n-i+1}\binom nk\binom{k-1}{i-1}$ is the degree in the symmetric group $S_n$ of the irreducible character which corresponds to the partition $(n-k,i,1^{k-i})$ (the latter entry meaning the cell $1$ repeated $k-i$ times), and a conjecture on which I am working will substantively advance if I get the full range of values coming from $X_1\cup X_2$ without having to worry about whether or not I have repetition by pulling something from $X_1$ then something from $X_2$.
The bonus question:
What I really want to be true is that $X_1$, $X_2$, and $X_3$ are pairwise disjoint, so a proof that covers all 3 pairs simultaneously would be the gold standard, but I would not at all be surprised if each $X_i,X_j$ pair had something peculiar to it that required individualized effort.  For instance, one reason $n\geq 8$ is required is that the sets $X_2$ and $X_3$ have respectively corresponding partitions $(5,2)$ and $(4,3)$ of $n=7$ which both yield the integer $14$; they are not disjoint.
Potentially relevant information:
I readily note that, in this indexing, both $X_1$ and $X_2$ have internal redundancy, i.e., if viewed as multisets, there would be multiple appearances of some integers.  I'm only asking about between sets, not within a given set.
My work:
Computer runs up into the $n=3200$ range say they are disjoint at least that far.
 A: This is to offer more evidence to the "computer run" by the OP.
Note that the members can be given by
$$X_1(n,k)=\binom{n-1}k \qquad \text{and} \qquad
X_2(n)=\frac{(n-1-k)(k-1)}{n-1}\binom{n}k.$$
It is not hard to show that if $p$ is a prime and $n=p^m$ then
$$\{(X_1(n,k) \mod n) \mod p: 1\leq k\leq n-1\}\equiv \{\pm 1 \mod p\}$$
while
$$\{(X_2(n,k) \mod n) \mod p: 2\leq k\leq n-2\}\equiv \{0 \mod p\}.$$
Hence for such numbers, $X_1$ and $X_2$ are disjoint.
A: Note that for a fixed n, one gets the same value in X_2 for both k and n-k, so most of these values occur twice (or more) in X_2. This also is observed in X_1. So if there is a common value between the two sets, we should find a $j$ with $2j \leq n-1$ and $k \lt j$ so that (n-1) choose j is a value for some $k$ in $X_2$. This leads to an interesting relation between $ n,k, $ and $j $ which shows very few large primes are involved.
(If a similar condition could be established for X_3, so that only $k \leq n/2$ need be considered, this would help with the analysis. )
Setting up the equation implied by a shared value, and using $k +1\lt j$ (which is justified by the above observations: the cases $k=j$ and $k+1 =j$ are easily handled separately), we can cross multiply and divide out by the common term $n-k-1$ to get the relation
$$(k-1)n(k+1)\cdots(j) = (n-1)(n-k)(n-k-2)\cdots(n-j).$$
I will rewrite this using $n=m+k$ and $j=k+d$ where $d \geq 2.$ The LHS is
$$m+k \cdot k-1 \cdot 1 \cdot k+1 \cdot k+2 \cdots k+d$$
And the RHS is
$$ m+k-1 \cdot 1 \cdot m \cdot 1 \cdot m-2 \cdots m-d.$$
I have left off parentheses and put in 1 to draw a correspondence between the two sides.
Note that if $d=1$, an analogous form and divisibility constraints would have $m+k$ dividing $m$, and $d=0$ is resolved even more simply. Note also that we have an approximate relation : $m^d$ is close to $k^{d+1}$. A more precise and messier formulation is possible, but this one should suggest that a tight numerical relationship exists given one of $k,m,d$.
We have more. Let $q$ be a factor of $m+k-1$. Suppose further that $q$ is coprime to $k+d$. Then $q$ is coprime to $m+k$, divides some sub product of terms on the LHS, and also divides the corresponding product of terms on the RHS. So $q^2$ divides both sides when it is coprime to $k+d$. A similar statement holds for $q$ dividing $m+k$ and coprime to $m$ ( and thus $k$ ).
One also observes from this that there can be very few large primes involved in the product: either they are in $ n-1 $ (and therefore in $ k+d$), or they are in $n$ and therefore in $k$, or they are in neither and the same constellation of large primes occurs "inside or near the dots". Note that all primes involved in the product must be less than $ k+d+1$, and those primes greater than d dividing $n(n-1)$ and not dividing $(k+d)k$ must occur with multiplicity 2 or greater in one of the terms.
I suspect these observations and one other (not yet observed) will lead to an impossibility proof for $d=2$. In any case, one can strongly restrict the search for $n$ and $k$ given a fixed $d.$
Gerhard "Got It By The Tail?" Paseman, 2019.08.27.
