How many five-digit positive integers can be formed by arranging the digits $1,1,2,3,4$ so that the two 1s are not next to each other?
Solution 1

Since the two 1s are not next to each other, then the two 1 s can be placed in the following pairs of positions, reading from the left: 1st and 3rd, 1st and 4th, 1st and 5th, 2nd and 4th, 2nd and 5th, 3rd and 5th.

There are 6 such pairs of positions.

Choose one of these pairs, say 1st and 3rd. This gives the number $1 \_1$

There are 3 digits left to place. We place these from left to right.

There are 3 possible digits that could go in the leftmost empty position.

After this digit is placed, there are 2 possible digits that could go in the next empty position. Finally, the remaining digit is placed in the remaining empty position.

This process works for each of the pairs of positions for the two 1 s.

Therefore, there are $6 \times 3 \times 2=36$ such five-digit integers ( 6 pairs of positions for the 1 s, 3 choices for the first empty position, and 2 choices for the second empty position).

## Solution 2

Since there are 5 digits to arrange and 2 of them are the same, there are $\frac{5 !}{2 !}=\frac{120}{2}=60$ arrangements:

To see this, replace one of the 1 s with and $X$ so that the "digits" were $1,2,3,4, X$. There would be 5 choices for the first digit, 4 choices for the second digit, and so on, giving 5 ! $=5 \times 4 \times 3 \times 2 \times 1=120$ arrangements of the digits.

If we now replace the $X$ with a 1, each arrangement is now counted 2 times. For example, $43 X 21$ and $4312 X$ become the same arrangement.

Therefore, we need to divide the total 120 by $2 !=2$ since each arrangement is double-counted.

In some of these arrangements, the 1s will be next to each other and in some they will not be. We will count the arrangements with the 1s next to each other and subtract this number from the total.

If the two 1s are together, we can imagine arranging the four objects $11,2,3,4$.

There are $4 !=24$ such arrangements.

Thus, there are $60-24=36$ arrangements of $1,1,2,3,4$ with the two 1 s not next to each other.

ANSWER: 36