In the question, we have to find the possible number of 6 digit numbers formed by the numbers 0, 1, 3, 5, 7, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

The first position from left will have five choices because zero cannot be assigned to that position because then our number will become a five digit number instead of six , the second position will also have five choices because when a number is occupied by the first position then four numbers are left but we ignored zero for the first position and for the second position we can use zero, the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form six digit numbers with the help of given data is 5 × 5 × 4 × 3 × 2 × 1 = 600

Numbers which are divisible by 10 should have zero at their one’s place , so we will fix zero on one’s place, and the rest of the positions will have choices accordingly.

The number of ways in which we can form six digit numbers when zero is fixed in one’s place and which are divisible by 10 is 5 × 4 × 3 × 2 × 1 × 1 = 5! = 120