combo + number theory !

Since $42=2\times3\times7$ , we need to find the numbers of $2$ 's, $3$ 's and $7$ 's (i.e. the index of them) in the prime factorization of $2007!$ , then find the least amongst the three. Trivially, the number of $7$ 's should be the least. This is achieved by $\displaystyle\left\lfloor\frac{2007}{7^1}\right\rfloor + \left\lfloor\frac{2007}{7^2}\right\rfloor + \left\lfloor\frac{2007}{7^3}\right\rfloor + \cdots=286+40+5=\fbox{331}$ , because the number of numbers divisible by $7$ under $2007$ is $286$ , then for each $49$ numbers we have an additional $7$ in the prime factorization of $2007!$ , and so on.