In how many ways can 2310 be expressed as a product of 3 factors ? ( This will not involve Combinatorics analysis !)
Hello, SheekhKebab!
In how many ways can 2310 be expressed as a product of 3 factors?
(This will not involve Combinatorics analysis!) . It won't?
I suppose a brute-force LIST is acceptable . . .
. . $\displaystyle \begin{array}{c}2\cdot3\cdot385 \\ 2\cdot5\cdot231 \\ 2\cdot7\cdot165 \\ 2\cdot11\cdot105 \\ 3\cdot5\cdot154 \\ 3\cdot7\cdot110 \\ 3\cdot11\cdot70 \\ 5\cdot7\cdot66 \\ 5\cdot11\cdot42 \\ 7\cdot11\cdot30 \end{array}$
Answer: 10 ways.
same as Soroban brute force list except I included 1 as a possible factor, and the first two factors don't have to be prime numbers
such as 6 11 35 and 7 10 33 are included in the list
There are 16 triplets of the form {1,y,z} since there are 32 divisors of 2310
I don't know how but I sense that there is a way to count the remaining 25. Is that what you are after?