If n is a natural number and n = \(p{^{x_1}_{1}}p{^{x_2}_{2}}p{^{x_3}_{3}}\), where p1, p2, p3 are distinct prime factors, then the number of prime factors for n is

1
x1 + x2 + x3
2
x1x2x3
3
(x+ 1)(x+ 1)(x3 + 1)
4
None of the above

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