Problem - 2479
How many different $6$-digit numbers can be formed by using digits $1$, $2$, and $3$, if no adjacent digits can be the same?
Answer
$96$
This problem can be solved by using the multiplication principle:
- First digit, no restriction: $3$ choices
- Second digit, can't be the same as the first one: $2$ choices
- Third digit, can't be the same as the second one: $2$ choices
- $\cdots$
Hence, the answer is $3\times 2\times 2\times 2\times 2\times 2=3\times 2^5=\boxed{96}$