You're looking for triples of three numbers (which I'm going to list in nondecreasing order) that sum to 36, and where each one is at least 5.
Instead you can look for triples of three numbers $p \le q \le r$ that sum to $36 - 15 = 21$, where each is nonnegative, and then the numbers you originally sought were $p+5, q+5, r+5$. (i.e., there's a 1-1 correspondence between the collection of partition-sizes that you seek and the collection of $(p, q, r)$ with $0 \le p \le q \le r$ and $p+q+r = 21$.)
These latter triples can be divided into those with $p = 0$, where the remaining numbers sum to 21; those where $p = 1$, and the remaining numbers sum to $20$...but both are at least $1$, those where $p = 2$, and the remaining numbers sum to $19$, but both are at least $2$, etc.
But the same trick applies to those: pairs of numbers, in nondecreasing order, summing to to $19$, but both being at least $2$ have the same count as pairs of (nondecreasing) numbers, summing to $15$, but both being at least $0$.
Let's write
$$
C(n, k)
$$
for the number of $n$-element sets of nonnegative numbers, in increasing order, that sum to exactly $k$. Then what I've said above shows that the number you're looking for is $C(3, 21)$, and that
$$
C(3, 21) = C(2, 21) + C(2, 21-3\cdot 1) + C(2, 21-3\cdot 2) + \ldots + C(2, 21-3\cdot 7)
$$
Now how large is
$C(2, k)$? It, too, satisfies a recurrence: the first number is either $0$ (in which case the remaining number must sum to 21), or it's 1, in which case the remaining number must be at least 1 and sum to 20, i.e., the count of which is the same as numbers that are at least 0 and sum to 19, etc.
$$
C(2, k) = C(1, k) + C(1, k-2 \cdot 1) + C(1, k-2 \cdot 2) + \ldots + C(1, k-2 \cdot (k/2))
$$
where the $k/2$ in the last term should be rounded down [i.e., if $k$ is odd, then we end with $C(1, 1)$ rather than $C(1, -1)$. ]
Now how large is $C(1, s)$ for any nonnegative $s$? It's exactly $1$.
That means that $C(2, k)$ is exactly $\lfloor \frac{k+1}{2} \rfloor$.
So
$$
C(3, 21) = \lfloor \frac{22}{2} \rfloor + \lfloor \frac{19}{2} \rfloor + \lfloor \frac{16}{2} \rfloor + \lfloor \frac{13}{2} \rfloor + \lfloor \frac{10}{2} \rfloor + \lfloor \frac{7}{2} \rfloor + \lfloor \frac{4}{2} \rfloor + \lfloor \frac{1}{2} \rfloor \\
= 11+9 + 8 + 7 + 5 + 3 + 2 = 45.
$$
This seems surprisingly low to me, but I don't see any obvious error (except that my "round down $(k+1)/2$" answer for $C(1, k)$ could be off by one), so I'm going to go ahead and submit it as an answer, and if not an answer, at least a suggested path for you to follow in getting to the correct answer.
Let me just sanity check by writing them down...there aren't that many.
12, 12, 12
11, 12, 13
11, 11, 14
10, 13, 13
10, 12, 14
10, 11, 15
10, 10, 16
9, 13, 14
9, 12, 15
9, 11, 16
9, 10, 17
9, 9, 18
8, 14, 14
8, 13, 15
8, 12, 16
8, 11, 17
8, 10, 18
8, 9, 19
8, 8, 20
7, 14, 15
7, 13, 16
7, 12, 17
7, 11, 18
7, 10, 19
7, 9, 20
7, 8, 21
7, 7, 22
6, 15, 15
6, 14, 16
6, 13, 17
6, 12, 18
6, 11, 19
6, 10, 20
6, 9, 21
6, 8, 22
6, 7, 23
6, 6, 24
Hunh. I seem to get 38. So I probably DID have an off-by-one error in my formula for $C(2, k)$. Anyhow, the answer is 38, and you can work out my off-by-one error by back-tracing, if it pleases you do to so.
Best Answer
Think of your elements as of $n$ balls which are to be placed into $k$ urns. We can place the balls in a row, and then insert $k-1$ dividors, separating thge balls into $k$ groups. Each group containing the balls which are to be placed into the corresponding urn. We have $n+k-1$ places for dividors and balls, and k-1 dividors to choose from them. So there are: $\binom{n+k-1}{k-1}$ possibilities.