Problem: how many 5 digit nos are there if the 2 leftmost digits are odd and the digit 4 ca n ' t appear more than once in the number ? could someone please provide a solution using a approach other than ( 1 - x ( none ) ) approach ?
Rationale: a . 4 is used once : oo * * 4 - - > ( 5 * 5 * 9 * 9 ) * 3 : 5 choices for the first digit as there are 5 odd numbers , 5 choices for the second digit for the same reason , 9 choices for one of the two * ( not - 4 digit ) , 9 choices for another * ( not - 4 digit ) , multiplied by 3 as 4 can take the place of any of the three last digits ( oo * * 4 , oo * 4 * , oo 4 * * ) ; b . 4 is not used : oo * * * - - > 5 * 5 * 9 * 9 * 9 : the same logic as above . 5 * 5 * 9 * 9 * 3 + 5 * 5 * 9 * 9 * 9 = 24300 . answer : a
annotated_formula: multiply(multiply(multiply(multiply(subtract(multiply(4, 5), 1), 5), 5), 5), const_10)