I did analysis of your problem in my dream last night, I read your question and during my sleep, I saw a dream in which I was analyzing the equation , so here is what I got

Since you already know the concept of Inclusion-Exclusion, lets consider the case

4 3 2 1

here given n = 4, a = 3, b = 2, c = 1

denotation of f(n) shows all the possible cases that can occur for all n = 1,2,3,…n

if n = 4, f(n) = 35, where various combinations will be(in sequence a,b,c)

```
0(combinations 1) = 0 0 0
1(combinations 3) = 1 0 0, 0 1 0, 0 0 1
2(combinations 6) = 2 0 0, 0 2 0, 0 0 2, 1 1 0, 1 0 1, 0 1 1
3(combinations 10) = 3 0 0, 0 3 0, 0 0 3, 2 1 0, 2 0 1, 1 2 0, 1 0 2, 0 1 2, 0 2 1, 1 1 1
4(combinations 15) = 4 0 0, 0 4 0, 0 0 4, 3 1 0, 3 0 1, 1 0 3, 1 3 0, 0 1 3, 0 3 1, 2 2 0, 2 0 2, 0 2 2, 2 1 1, 1 2 1, 1 1 2
```

but as asked in the question, we have to restrict a, b, c with 3, 2, 1 respectively, for that we exclude all the undesired cases which can be done easily with Inclusion-Exclusion Principle

```
f(n-a-1) = f(4-3-1) = f(0) = 1 (combinations (4 0 0))
f(n-b-1) = f(4-2-1) = f(1) = 4 (combinations (0 3 0, 0 4 0, 1 3 0, 0 3 1))
f(n-c-1) = f(4-1-1) = f(2) = 10 (combinations (0 0 2, 0 0 3, 1 0 2, 0 1 2, 0 0 4, 1 0 3, 0 1 3, 2 0 2, 0 2 2, 1 1 2))
f(n-a-b-2) = f(4-3-2-2) = f(-3) = 0
f(n-b-c-2) = f(4-2-1-2) = f(-1) = 0
f(n-a-c-2) = f(4-3-1-2) = f(-2) = 0
f(n-a-b-c-3) = f(4-3-2-1-3) = f(-5) = 0
```

so, by the Principle of Inclusion we get

`ct = 35 - 1 - 4 - 10 = 20`