How many different ways can 6 numbers be chosen from a set of 49, without regard to order?

Choosing $6$ numbers without regard to order is a combination: $\binom{49}{6}=\frac{49!}{6!\,43!}=\frac{49\cdot48\cdot47\cdot46\cdot45\cdot44}{720}=13{,}983{,}816$, choice B. Choice C ($49^6$) counts ordered choices with repetition allowed, so it is far too large. Choice D ($6^{49}$) swaps the base and the exponent. Choice A ($294$) is $49\cdot6$, a product that does not model any selection. Because the order of the numbers does not matter, we divide the permutations by $6!$, the number of ways to order six numbers.