arjunbt wrote:
Can we do it like this ?
_ _ _
2 Possibilities for 100th digit 8 and 9,
9 Possibilities for 10th Digit,
8 Possibilities for Unit digit.
that gives us 2 * 9 * 8 = 144,
Since we need only odd numbers 144/2 = 72.
And thats the right answer.
Can anyone tel me if i am missing something or this method is alright ?
Yes, you are missing something and there is a reason why Ian did what he did.
Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different?
1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72
How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4.
[highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight]
Number of even numbers = 72 - 32 = 40
Now take numbers in the range 800 - 900.
Total numbers where all digits are different = 72 (as before)
[highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit)
Number of even numbers = 72 - 40 = 32
So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).
The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.
If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.
I am definitely taking more than the prescribed 2 minutes to solve these questions. Am i missing something ?