how many different 3 - digit multiples of 5 can be composed of digits 2 , 7 , 4 , and 5 if none of the digits is repeated ?
= 3 c 2 x 2 = 6 can you explain your method . i solved it differently xyz , where z must be 5 . therefore 1 variation of digit in z . y can be any of the 3 possible choices . x can be any of the 2 possible choices . 2 + 3 + 1 = 6 b
divide(subtract(multiply(7, 4), 4), 4)