Derek’s phone number, 336-7624, has the property that the three-digit prefix, 336, equals the product of the last four digits, 7 × 6 × 2 × 4. How many seven-digit phone numbers beginning with 336 have this property?
So I went number by number and found the number of combination including each of the numbers.
For when the four digits include 1 the only way is for the numbers to be in any combination of 1,6,7,8. And there are 4! arrangements
For when the four digits include 2, there are two ways (without using 1). Either 2,4,6,7 or 2,3,7,8, and each has 4! arrangements totaling to 2(4!) arrangements.
For when the four digits include 3, there is one way (not including 2, or 1), the numbers 3,4,4,7. There are 4!/2 ways to arrange those.
There are no more ways to make 336 without using 1,2, or 3.
Adding all the cases ends up with 84.
This is the correct answer, but my question is, what is the fastest way to do this, or is there no other way?
I don't know a faster way. You didn't say how you found the four digit possibilities. If you factor $336=2^4\cdot 3 \cdot 7$ you can find the four digit possibilities rather easily. One digit has to be $7$. One has to be $3$ or $6$. Then you split the remaining $3$ or $4$ factors of $2$ among two digits.