How many odd 4 digit numbers can be formed from the digits 0 through 7 if there must be a 4 in the number and numbers can't repeat?

The provided answer is 320... I don't know how to get that.

  • $\begingroup$ The last digit can be one of $1,3,5,7$. If the first digit is $4$, how many options do we have to fill the other digits? If it is the second or the third? Remember that the first digit can't be zero(I assume). $\endgroup$ – chubakueno Mar 23 '14 at 0:10
  • $\begingroup$ Anything in particular you've tried? $\endgroup$ – MCT Mar 23 '14 at 0:12
  • $\begingroup$ just like the first comment stated, that is where i am at. I don't know what to do next... nothing brings me to this 320. $\endgroup$ – Danny Watts Mar 23 '14 at 0:14
  • 1
    $\begingroup$ If the first digit is $4$: We have "used" one number from $\{1,3,5,7\}$, and we have also "used" the $4$, therefore we have $6$ remaining numbers to choose from to fill the second digit, and $5$ to fill the third. There are $4\times6\times5$ ways to do this. The reasoning for the case where the second digit is $4$ is similar, we have $4\times5\times5$ ways of doing that(since we can't choose $0$ for the first one). Can you finish? $\endgroup$ – chubakueno Mar 23 '14 at 0:21

Your number must be odd, so the last digit must be one of $\{1, 3, 5, 7\}$ Now that we've got that dealt with, we can just consider the first three digits.

Our three digit number will be of three possible forms: $4 ? ?, ? 4 ?, ? ? 4$

Case 1: $4??$ - We have now used two digits (the four and whatever odd number we pick to be the last digit), so there are $6 * 5$ ways to pick the next two digits without replacement.

Case 2: $?4?$ - It's similar to the last case except for one fact: The leading digit cannot be $0$ (or else it would be a three digit number). So, there are $6 - 1 = 5$ ways to pick the first digit, and $5$ ways to pick the third. The same holds for $??4$, so that's $2 * 5 * 5$ counting both.

Finally, we have four possibilities for our last digit. To recap, in total we have $4(6*5 + 2*5*5) = 320$ ways to do this.

Problem to apply what you've learned from this post

How many even $5$ digit numbers can be formed from the digits $1$ through $9$ given that there must be a $7$? Note that the digits need not be distinct.

  • $\begingroup$ Are there 7*6*5*4*4 = 3360 possibilities? $\endgroup$ – Danny Watts Mar 23 '14 at 0:37

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