We assume $k$ is a fixed small integer.
For $k\le 2$ , there are no solutions :
The case $k=1$ is obvious since $n$ must be prime.
If $k=2$, then $n$ and $n-2+1=n-1$ have both at least $2$ prime factors (counting multiplicity), giving at least $2\cdot2-1=4-1=3$ prime factors (counting multiplicity) for $\binom{n}{2}$, a contradiction.
Hence we assume $k\ge3$.
If $k=3$, then $n$, $n-1$ and $n-2$ must have together 5 prime factors (counting multiplicity), so $n$ and $n-2$ must have two prime factors and n-1 must be prime, which is only possible for $n=6$ (and allowing negative integers, also $n=-4$), since $n$ is even and $\binom{2n}{3}$ must be 3-almost prime. This is four times the $n-1$-st square pyramidal number, so the $n-1$-th square pyramidal number must have $3-2=1$ prime factors, making it prime, and the only square pyramidal number which is prime is 5.
For $k=4$, the largest example is $n=15$. Proof:
Case 1: $n$ mod $3\ne0$.
Then $n$ and $n-3$ are both composite, and at least one of them is odd. So $(n-1)\cdot(n-2)$ has at most $8-4=4$ prime factors (counting multiplicity). Since in this case $\binom{n}{4}=(n/GCD(n,4))\cdot((n-1)/GCD(n-1,12))\cdot((n-2)/GCD(n-2,12))\cdot((n-3)/GCD(n-3,4))$ and at least one of $n$ and $n-3$ is odd an composite and $4=2^2$ has no odd prime factors, it follows that at least one of the factors is $1$. Since the smallest odd composite not divisible by $3$ is $25$ and $14$ is the largest $n$ such that at least one of the factors of $\binom{n}{4}$ is $1$ equals $14$, we arrive at a contradiction by checking $n\le7$.
Case 2: $n$ mod $3=0$.
Then $\binom{n}{4}=(n/GCD(n,12))\cdot((n-1)/GCD(n-1,4))\cdot((n-2)/GCD(n-2,4))\cdot((n-3)/GCD(n-3,12))$, which has at least four prime factors for $n\gt15$. Checking the remaining cases where $n$ and $n-3$ are both not equal to 3, we get $n=9,12,15$ (or if negative $n$'s are allowed, also $n=-6,-9,-12$).
Q.E.D.
If $k$ is a prime power, we can use similar methods to find the largest such $n$. However, if $k$ isn't a prime power, no such methods are applicable: Dickson's conjecture implies that there are infinitely many such $n$'s. For example, for $k=6$ the largest solutions up to million are $n=185534, 670767, 709694$, while for $k=5$, the largest solution is $n=62$, but we need a calculator to prove this easily.
It can be proved by using that if $k$ is a prime power, then at least one $a$ in the interval $[n-k+1,n]$ is a divisor of $LCM(1..k)$, so we need to check only up to $n=LCM(1..k)+k-1$.
Let $a(k)$ be the largest (known) integer $n$ such that $\binom{n}{k}$ has $k$ prime factors (counted with multiplicity) and such that $n$ and $n-k+1$ are both composite.
If $k$ is the prime power, then at least one of $a(k)-n+1..a(k)$ is a divisor of $LCM(1..k)$, making $a(k)$ finite.
If $k$ isn't a prime power, then it can be proved that Dickson's conjecture implies that $a(k)$ is infinite, so searching for the large $n$ such that both $n$ and $n-k+1$ are composite and such that $\binom{n}{k}$ has exactly $k$ prime factors (counted with multiplicity) makes sense. The lower bound for such largest known $n$'s in this case are in parentheses.
Then $a(1..18)$ are the following:
$$?, ?, 6, 15, 62, \infty (\ge833106029305094), ..., 63, 143, 185, \infty (\ge1387429387915), 319, \infty (\ge 319), 238, \infty (\ge 319), \infty (\ge 799), 1241, 341, \infty (\ge 923)$$.
Main task: Extend this list when $k$ is a prime power.
Note: The parenthetical results for $k\in\{6,10\}$ can be easily improved.