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(D) The total number of $5$-digit numbers that can be formed using the digits $1$ and $2$ is $2^5 = 32$.The numbers where all digits are the same are

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(D) The total number of $5$-digit numbers that can be formed using the digits $1$ and $2$ is $2^5 = 32$.The numbers where all digits are the same are $(1, 1, 1, 1, 1)$ and $(2, 2, 2, 2, 2)$.We are looking for numbers where exactly one digit is different. This means $4$ digits are the same and $1$ digit is different.Case $1$: Four $1$s and one $2$. The number of arrangements is $\frac{5!}{4!1!} = 5$.Case $2$: Four $2$s and one $1$. The number of arrangements is $\frac{5!}{4!1!} = 5$.Total such numbers = $5 + 5 = 10$.

How many $5$-digit numbers can be formed using the digits $1$ and $2$,such that exactly one digit is different from the others?

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