CircularArrangement Basic

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?


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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 $3\times 2\times 2\times 2\times 2\times 2=3\times 2^5=\boxed{96}$

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