# Math

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How many odd four-digit numbers, all of the digit different can be formed from the digits 0 to 7, if there must be a 4 in the number?

There are four options for the last digit (1, 3, 5, or 7). One of the other three digits only has 1 option (for the 4). One of the other remaining digits will have 6 options and the final digit will have 5 options (because they can't repeat the first two) Therefore, the answer is:

4*1*5*6=120

I do not have much confidence in my answer. Could someone please tell me if I did this right? Thanks for your help.

• Math -

The "4" could be in
the first place,
the 2nd place, or
the 3rd place

so you could have 4 _ _ (odd)
or
_ 4 _ (odd)
or
_ _ 4 (odd)

the first of these could be done
1 x 5 x 4 x 4 = 80
the 2nd
5 x 1 x 4 x 4 = 80
the 3rd
5 x 4 x 1 x 4 = 80

for a total of 240

explanation of the first case:
1. fill in the 4 possiblities to make it odd, at the far right
2. fill in the "4" at the front, 1 way only
3. two digits have now been used up, leaving 5 of the remaining to go in the second spot.
4. three digits have now been used, leaving 4 of the remaining to go in the third spot.
thus : 1x5x4x4

Many of these kind of questions can be easily done by splitting them up into "cases"

• Math -

Reiny,

Thanks but I'm stumped on part 3 of the explanation for the first case. We have used up 2 digits when we move on to the second space but didn't we start with 8 digits (0-7) so I don't understand why we aren't left with 6 choices for the second spot (8-2). Please help me understand this.

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