How many different sets of positive integers $(a, b, c, d)$ are there such that $a \lt b \lt c \lt d$ and $a + b + c + d = 41$? Is there a general formula which I can use to calculate this and if it's with proof or reasoning would be great as well. Even if you could please solve this $4$-variable ordered set of positive integers case would be a great help indeed. Sorry I am weak in this topic.
Thanks Robert for your pointers in modifying it to
$(4\times1) + (3\times2) + (2\times3) + (1\times4) = 41$, where there are no additional constraints on $x_i$
To solve this must do it manually by cases or is there a short way of calculating it using "Bars and Stars" (of which I am aware of the method)? The answer given is $321$.
But when I calculate it like this it doesn't seem to match the answer:
imagine the new given equation as containing $(4+3+2+1) = 10$$x_i$'s individually so as equivalent to finding the number of positive integer solutions for $b_1 + b_2 + ... + b_{10} = 41$ which is $C(41 - 1, 10 - 1) = C(40, 9)$. Now since we have $4x_1$'s that are identical, we need to divide this by $4!$ (for its permutations of "overcounting") and similarly divide next by $3!$ (for $3x_2$'s that are equal), and $2!$ for the $2x_3$'s that are equal. How is my approach wrong here? anybody, please help me? Thanks again
As per Robert's transformation with Anil's suggestion of the generating functions method, here's my work on it:
I actually used an online calculator called "Sage Math" to do this via this website:
https://www.whitman.edu/mathematics/cgt_online/book/section03.01.html
With these modifications
$f(x)=(1-x^4)^{-1}\times(1-x^3)^{-1}\times(1-x^2)^{-1}\times(1-x)^{-1}$
show(taylor($f,x,0,33$))
And have gotten the verification  that $321$ is indeed the answer after some tedious algebra (please note that since we are looking for the coefficient of $x^41$ in the expansion Anil wrote which has $x^(4+3+2+1) = x^10$ factored outside meaning in the remaining Newton's Binomial Expansions with negative exponent, we only need to determine the coefficient of $x^(41-10)= x^31$ in it which is "$321$" as the answer suggested (please see image below):


Thanks some info from OEIS, and some help from Mathworld, we have the following:
Let the number of partitions of $n$ into $k$ distinct parts be $Q(n, k)$. Then $Q(n,k)$ has the recurrence:
$$Q(n,k) = Q(n-k, k) + Q(n-k, k-1)$$
where $Q(n,1) = 1, Q(n,0) = 0$.

If you prefer, we have a (possibly easier?) method: $Q(n, k) = P(n- \binom{k}{2}, k)$, where $P(n,k)$ has the recurrence:
$$P(n,k) = P(n-1, k-1) + P(n-k, k)$$
where $P(n,1) = P(n,n) = 1$. We'll want $P(35,4)$ of course. Either way, the OEIS link gives $Q(41,4) = P(35,4) = 321$.

If you must have a closed-form function, Mathworld has you... "covered," in the loosest sense of the term, and only for $P(n,k)$ up to $k=4$. The function involved is
$$P(n,4) = \textstyle \frac1{864} \displaystyle \left\{ 3 \left(n+1 \right) \left[2n \left(n+2 \right)-13+9 \left(-1 \right)^n \right] - 96 \cos \frac{2 \pi n}{3}+ \\ \left( 108 \left(-1 \right)^ \left(n/2 \right) \right) \big(\bmod \left(n+1,2 \right) \big)+32 \sqrt{3} \sin \left(\frac{2 \pi n}{3} \right)\right\}$$
And yes, for $n=35$ it does in fact evaluate to $321$.
