How Many Odd 4-Digit Numbers with a 4 Can Be Formed from Digits 0-7?

[kerrwilk]

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.

Homework Equations

The Attempt at a Solution

First way (the "4" in the first spot) 1X6X5X4=120
2nd way ("4" in second spot) 5X1X4X4=80
3rd way ("4 in third spot) 5X4X1X4=80 for a total of 280 ways

 

[gb7nash]

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