$x,y,z\text{ are natural numbers, not greater than }6.\text{ How many ordered triples }(x,y,z)\text{ are }$

$\text{ there such that the product xyz is divisible by }10\text{ ? }$

$\text{ Total }{6}^3=216$

$5^3=125\text{ product do not have }5\text{ as a factor }(1,2,3,4,6)$

$3^3=27\text{ product do not have }2\text{ as a factor }(1,3,5)$

$2^3=8\text{ product do not have }2\text{ \& }5\text{ as factors }(1,3)$

$\therefore 216-125-27+8=72$