kerrwilk said:
Homework Statement
Determine how many odd 4-digit numbers with all of the digits different can be made from the digits 0 to 7 if there must be a 4 in the number.
This might be an easier way of looking at the problem. So if I understand correctly, we want to find all 4-digit numbers such that a 4 resides in the first, second, or third slot, a member from {1,3,5,7} resides in the last slot, and members from {0,1,2,3,4,5,6,7} remain in the remaining two slots where all 4 digits are distinct.
So 0143 and 5427 are examples of valid numbers.
So think of there as being 4 slots, so we have:
_ _ _ _
First, choose a slot for the 4 to reside in. There are 3 ways of doing this. (Do you know why?)
So now that we have chosen a slot for the 4, choose an odd number to reside in the last slot (since the 4-digit number itself must be odd). There are 4 odd numbers between 0 and 7: 1,3,5,7. So there are 4 ways of doing this.
Now, let's choose a number to reside in the second slot. Remember, at this stage there are only 6 numbers left to choose from between 0 and 7, since we already placed the 4 and we already placed an odd number in {0,1,2,..,7}. So there are 6 ways to do this.
Lastly, place a number in the 3rd slot. There are 5 numbers to choose from.
So by the multiplication principle, we get: 3*4*6*5 = 360 possibilities