Question 1111294: Five digit numbers are made using the numbers 0 to 7. Determine the number of odd numbers that can be made if each digit can only be used once.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
In addition to the requirement that digits not be repeated, the only restriction is that the number must be odd -- that is, the units digit must be odd.
So choose the units digit first; you have 4 choices.
After that, you can choose the digits in any order you want. You started with 8 different digits, so you have 7 choices for the 2nd digit you choose, the 6 choices for the 3rd, and so on.
The total number of different 5-digit numbers with the given restrictions is
4*7*6*5*4 = 3360
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
If it is allowed in this problem for five digit numbers to start from the zero digit, then the solution by other tutor is good.
But it is commonly assumed in the everyday life that, when consider "n-digit numbers", then the first digit is not "0".
Under this constraint, the solution of the previous post must be changed in this way:
1. Choose the "units" digit by any way among 1, 3, 5 or 7.
You will have an odd number, and you have 4 options at this step.
2. As the left-most, you can choose any of remaining 8-1 = 7 digits EXCEPT zero.
So you have (8-1)-1 = 6 options in all at this step.
3. For 2-nd position from the left, you have 8-2 = 6 options among remaining digits.
For 3-rd position from the left, you have 5 options among remaining digits.
For 4-th position from the left, you have 4 options among remaining digits.
4. In all, you have 4*6*6*5*4 = 2880 choices and 2880 of 5-digit different odd numbers.
|
|
|
