How many six-digit odd numbers, greater than 6,00,000, can be formed from the digits 5, 6, 7, 8, 9, and 0 if repetition of digits is allowed repetition of digits is not allowed.

Answer

240

Explanation

We have 6 digits, viz., 5,6,7,8,9, and 0 and we have to form numbers (integers) greater than 6,00,000, which are odd.
So the first place (lakh's position) should be `ge 6` and the last position (i.e., unit) must be odd, i.e., 5,7, or 9.
(a) When repetitions is allowed.
First place can be filled by 6,7,8, or 9 i.e., in 4 ways, last place can be filled by 5,7, or 9 i.e., in 3 ways and each of the remaining 4 places (i.e., 2nd, 3rd, 4th, 5th) can be filled by any of the 6 digits in 6 ways. Hence, the total numbers will be `4xx6xx6xx6xx6xx3=15552`.

The total number of numbers that can be formed is
`4xx6xx6xx6xx6xx3=15552`.
(b) When repetitions is not allowed.
Since we have restrictions on first and last places and no digit can be repeated, we have the following algorithm: 1st place having numbers `ge 6` and last place having odd numbers.

Thus, the first and the last place can be filled in 10 ways, and the remaining 4 places can be filled by the remaining 4 digits in 4!=24 ways. Hence, the total number of numbers that can be formed is `10xx24=240`.