Consider set of four non negative integers $S=\{s_1,s_2,s_3,s_4:s_i\in\mathbb{Z^+\cup\{0}\}\}$. Then there are the following possible broad cases:
$1$. All four integers are equal.
$2$. These integers are pairwise distinct.
$3$. There are three distinct integers.
$4$. There are two distinct integers appearing twice each.
$5$. There are two distinct integers with one appearing three times.
Question: What are all the possible cases with these four integers in $S$ considering the above broad cases?
Case $1$. $s_1=s_2=s_3=s_4$
Case $2$. $s_i\neq s_j$ for all $i\neq j$
Case $3$. a. $s_1\neq s_2$, $s_1\neq s_3$, $s_4= s_1$.
b. $s_1\neq s_2$, $s_1\neq s_3$, $s_4= s_2$.
c. $s_1\neq s_2$, $s_1\neq s_3$, $s_4= s_3$.
d. $s_2\neq s_3$, $s_2\neq s_4$, $s_1= s_2$.
e. $s_2\neq s_3$, $s_2\neq s_4$, $s_1= s_3$.
f. $s_2\neq s_3$, $s_2\neq s_4$, $s_1= s_4$.
g. $s_3\neq s_4$, $s_3\neq s_1$, $s_1= s_2$.
h. $s_3\neq s_4$, $s_3\neq s_1$, $s_2= s_3$.
i. $s_3\neq s_4$, $s_3\neq s_1$, $s_4= s_2$.
Case $4$. a. $s_1=s_2$, $s_3=s_4$, $s_3\neq s_1$.
b. $s_1=s_3$, $s_2=s_4$, $s_4\neq s_3$.
c. $s_1=s_4$, $s_2=s_3$, $s_4\neq s_3$.
Case $5$. a. $s_1=s_2=s_3$, $s_4\neq s_1$.
b. $s_2=s_3=s_4$, $s_2\neq s_1$.
c. $s_3=s_4=s_1$, $s_2\neq s_1$.
d. $s_4=s_1=s_2$, $s_3\neq s_1$.
I think there should be $24$ subcases in total and in the above there are only $18$ subcases.
Any help will be highly appreciated. Thank you!
Your approach is good! You just double-counted some subcases for Case 3. For example, note that subcase 3a is really the same as 3f: in both cases you have $s_1$ and $s_4$ as the same number, while $s_2$ is a second distinct number, and $s_3$ a third distinct number yet. To further see that these two subcases are really the same, it is helpful to indicate all identities and between the non-identities between the $4$ integers, as in both cases we have:
$s_1\neq s_2$, $s_1\neq s_3$, $s_1= s_4$, $s_2\neq s_3$, $s_2\neq s_4$, $s_3 \neq s_4$.
This notation also suggests a way to correctly count the number of subcases for Case 3: out of the $6$ identity/non-identity statements, there is exactly one identity statement, and 5 non-identity statements. So, you just need to find one pair of integers out of the $4$: once you have done so, your subcase is completely fixed: the pair of integers form an identity claim, while for the others you have all non-identity claims.
Doing this, you get:
A. $s_1 = s_2$, $s_1\neq s_3$, $s_1 \neq s_4$, $s_2\neq s_3$, $s_2\neq s_4$, $s_3 \neq s_4$. (your 3d and 3g)
B. $s_1\neq s_2$, $s_1= s_3$, $s_1 \neq s_4$, $bs_2\neq s_3$, $s_2\neq s_4$, $s_3 \neq s_4$. (your 3e)
C. $s_1 \neq s_2$, $s_1\neq s_3$, $s_1 = s_4$, $s_2\neq s_3$, $s_2\neq s_4$, $s_3 \neq s_4$. (your 3a and 3f)
D. $s_1\neq s_2$, $s_1\neq s_3$, $s_1 \neq s_4$, $s_2 = s_3$, $s_2\neq s_4$, $s_3 \neq s_4$. (your 3h)
E. $s_1 \neq s_2$, $s_1\neq s_3$, $s_1 \neq s_4$, $s_2\neq s_3$, $s_2 = s_4$, $s_3 \neq s_4$. (your 3b and 3i)
F. $s_1\neq s_2$, $s_1\neq s_3$, $s_1 \neq s_4$, $s_2\neq s_3$, $s_2\neq s_4$, $s_3 = s_4$. (your 3c)
Now adding up all subcases gets you to 15.