Question

How many five-digits numbers can be made having exactly two identical digits?

Answer 1

Consider $2$ cases:

Case $1:$

One of the identical digits is the first digit of the number:

There are $4$ options for the position of the other identical digit.

For each of these there are

$9$ options for the value of the identical digits (can't be $0$)

$9$ options for the first of the non-identical digits (can't be the same as the two identical digits)

$8$ options for the second of the non-identical digits, (can't be the same as the identical digits or the first non-identical digit)

$7$ options for the last of the non-identical digits.

So there are $4\times 9\times 9\times 8\times 7=4\times 4536$ numbers where the first digit is one of the identical digits.

Case $2:$The first digit is not one of the identical digits

There are ${}^4{C}_2=\frac{4!}{2!2!}=\frac{4\times 3\times 2}{2\times 2}=6$ ways to choose the position of the 2 identical digits. ($4$ since you can't choose the first digit)

For each of these there are

$9$ options for the first digit (can't be $0$)

$9$ options for the second non-identical digit (can't be the same as the first digit, but can be $0$)

$8$ options for the last non-identical digit (can't be the same as the first or second digit, can be $0$)

$7$ options for the value of the two identical digits (can't be the same as any of the other digits, can be $0$)

So there are $6\times 9\times 9\times 8\times 7=6\times 4536$ where the first digit is not one of the identical digits.

Hence there are$=6\times 4536+4\times 4536=10\times 4536$

$=45360$ five digit numbers with the required property.