0
$\begingroup$

I had the following problem:

We have the word BORBOTTIO. Find:

  1. All the anagrams
  2. All the anagrams that starts with BB
  3. All the anagrams where the same letters are close together.

The first two questions are very easy. What about the third? I think $5!$ but in the book, the solution is $36$.

Could someone help me?

$\endgroup$
  • 2
    $\begingroup$ There are $5!=120$ sequences where all of BB, TT, and OOO occur, and I can't imagine how these letters could be any closer together. I also can't imagine where the book gets $36$ $\endgroup$ – saulspatz Aug 22 at 19:28
1
$\begingroup$

The only way to get the result is that the groups of $B´s, T´s$ and $O´s$ are ordered consecutively. Thus we have the block $\textbf B_2\textbf T_2\textbf O_3$ and the letters $R$ and $I$, where $B_2=BB, T_2=TT, O_3=OOO$

We can arrange $\textbf B_2,\textbf T_2,\textbf O_3$ in $3!=6$ ways. And $\textbf B_2\textbf T_2\textbf O_3,R,I$ can be arranged in $3!=6$ ways as well. So the answer is $6\cdot 6=36$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.