[" 27.A five digit number divisible by "3" is to be "],[" formed by using the numerals "0,1,2,3,4" and "],[5" without repetition.The total number of "],[" ways in which can be done is "]

A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is

A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is

A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is

A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is

A five digit number divisible by 3 is to be formed using the numbers 0,1,2,3,4 and 5 without repetitions . The total number of ways in which this can be done is

A five digit number divisible by 3 is to be formed using the numbers 0,1,2,3,4 and 5 without repetitions . The total number of ways this can be done is :

A five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 und 5 without repetition. The total number of ways this can be done is

A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4, & 5 without repetition . if the total number of ways in which this can be done is n^3 , then (n!)/144 must be...