In how many possible ways can we write 3240 as a product of three positive…

Concept: To count ordered triples (a, b, c) of positive integers with a x b x c = N, write N in prime-factorised form and treat each prime's exponent separately. Distributing e identical copies of a prime among 3 distinguishable factors, allowing any factor to get zero, is a stars-and-bars count: C(e + 3 - 1, 3 - 1) = C(e + 2, 2) ways. Since the primes act independently, the total number of ordered triples equals the product of this count over every prime in N.

Application: Prime-factorise 3240:

3240 = 23 × 34 × 51

1. Distribute the three 2's among a, b, c: C(3 + 2, 2) = C(5, 2) = 10 ways.

2. Distribute the four 3's among a, b, c: C(4 + 2, 2) = C(6, 2) = 15 ways.

3. Distribute the one 5 among a, b, c: C(1 + 2, 2) = C(3, 2) = 3 ways.

4. Multiply the three independent counts, since the primes act independently: 10 x 15 x 3 = 450.

Cross-check: each factor above counts non-negative integer solutions (x, y, z) to x + y + z = e, the exponent of that prime going to a, b and c respectively. A triple (a, b, c) is uniquely determined once we fix, for every prime, how its exponent splits across the three factors, so this product enumerates every ordered triple with a.b.c = 3240 exactly once, confirming the total of 450.