# How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7 and 9 when no digit is repeated?

Question:

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7 and 9 when no digit is repeated? How many of them are divisible by 10?

Solution:

The first digit of the number cannot be zero. Thus, it can be filled in 5 ways.

The number of ways of filling the second digit = 5        (as the repetition of digits is not allowed)

The number of ways of filling the third digit = 4

The number of ways of filling the fourth digit = 3

The number of ways of filling the fifth digit = 2

The number of ways of filling the sixth digit = 1

$\therefore$ Required numbers $=5 \times 5 \times 4 \times 3 \times 2 \times 1=600$

For the number to be divisible by 10, the sixth digit has to be zero.

Now, the first digit can be filled in 5 ways.

Number of ways of filling the second digit = 4

Number of ways of filling the third digit = 3

Number of ways of filling the fourth digit = 2

Number of ways of filling the fifth digit = 1

Number of ways of filling the sixth digit = 1

Total numbers divisible by $10=5 \times 4 \times 3 \times 2 \times 1 \times 1=120$