Choosing one of each letter from a string of repeated "ABCD"s such that it is in order of "ABCD"

View the problem as

$$(A_1B_1C_1D_1) \_(A_2B_2C_2D_2) \_(A_3B_3C_3D_3) \_\ldots (A_nB_nC_nD_n) \_$$

Let's view the underscores as boxes. We want to allocate $4$ balls into those boxes. As we read from the left to the right, the first ball that we encounter will tell us which $A$ to pick, if it is at the $i$-th box, pick $A_i$. Similarly for the other balls.

The number of such allocation of $k$ balls to $n$ distinguishable boxes is $\binom{n+k-1}{k}$. Here is a relevant link.

In our context, $k=4$, hence the formula.

$$\binom{n+3}{4}=\frac{(n+3)(n+2)(n+1)n}{4!}$$

Equivalently, you are counting the number of $4$-tuples $(a,b,c,d)$ where $a,b,c,d\in\{1,\dots,n\}$ and $a\leq b\leq c\leq d$. Here the tuple $(a,b,c,d)$ means that you pick the $a$th $A$, the $b$th $B$, and so on.

If you instead had strict inequalities $a<b<c<d$ then there would be a very easy answer: for any set of four distinct elements of $\{1,\dots,n\}$, there is a unique way to put them in order to get an increasing $4$-tuple, so there are $\binom{n}{4}$ possibilities. With nonstrict inequalities, here is a trick to get a similar answer. Let us add three new symbols $R_2,R_3,R_4$ to our set $\{1,\dots,n\}$. Then I claim there is a bijection between the set of $4$-tuples $(a,b,c,d)\in\{1,\dots,n\}^4$ such that $a\leq b\leq c\leq d$ and the set of $4$-element subsets of $\{1,\dots,n,R_2,R_3,R_4\}$. Namely, given a $4$-tuple $(a,b,c,d)$, map it to the set $\{a,b,c,d\}$, except that if the $i$th element is repeated, you replace it with $R_i$. So for instance, a tuple with $a=b<c=d$ would map to $\{a,R_2,c,R_4\}$, with $b$ and $d$ getting replaced by $R_2$ and $R_4$. Conversely, starting with a $4$-element subset of $\{1,\dots,n,R_2,R_3,R_4\}$, you get a tuple by just put the numbers in increasing order and repeating numbers in the $i$th spot for any $R_i$ that is in your set. I'll leave it to you to verify that these operations are inverse to each other.

So, the number of $4$-tuples $(a,b,c,d)\in\{1,\dots,n\}^4$ with $a\leq b\leq c\leq d$ is $\binom{n+3}{4}$. More generally, a similar argument shows the number of nonstrictly increasing $k$-tuples of elements of $\{1,\dots,n\}$ is $\binom{n+k-1}{k}$.