coolbreeze2011 wrote:Can someone please show me how to solve this problem..
How many odd three-digit integers greater than 800 are there such that all their digits are different?
40
56
72
81
104
Let us see the integers between 800 to 900 first.
Units place can be any digit from 1, 3, 5, 7, 9, which means units digit can be filled in
5 ways.
Tens place can be filled with digits 0 to 9, but the digit which has already been used in the units place, cannot be repeated. Also the hundreds digit is fixed as 8, so it can be filled in
8 ways.
Hundreds place has to digit 8, which means
1 way.
Now let us see the integers from 900 to 999.
Units place can be any digit from 1, 3, 5, 7 (note that 9 will be fixed as the hundreds digit so it cannot be used in the units place, as digits should be different), which means units digit can be filled in
4 ways.
Tens place can be filled with digits 0 to 9, but the digit which has already been used in the units place, cannot be repeated. Also hundreds digit is fixed, so it can be filled in
8 ways.
Hundreds place has to digit 9, which means
1 way.
Therefore, required number of odd three-digit integers = (5 * 8 * 1) + (4 * 8 * 1) = 40 + 32 =
72
The correct answer is C.
