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Suppose we are given that

$$K = \{a,b,c,d,e,f\}$$

In how many 4th permutations of given subset $a$ and $b$ are adjacent?

So we have four choices for remaining letters when we put $a$ and $b$

$$ab\_\space \_$$

Now let us pick 2 letters out of 4 letters, which can be done in $\binom{4}{2}$ ways. However, $a$ and $b$ have to be together so we can consider them as an object yielding $3!$ Thereby, we finally have that

$$\binom{4}{2}3! = 6^2 = 36$$

Am I right?

Regards

Fiv
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  • Close., I would consider $a$ and $b$ to be adjacent in the string $baef$ as well. Your answer would be correct if not only $a$ and $b$ needed to be adjacent but also that $a$ must occur before $b$ so you missed counting half of them. – JMoravitz Nov 07 '18 at 17:23
  • @JMoravitz Could you be more clear with what you mean? – Fiv Nov 07 '18 at 17:39
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    Not really... I can try again. You counted strings like $\overline{ab}cd, \overline{ab}ce, \overline{ab}cf, \overline{ab}dc, \dots, c\overline{ab}d, c\overline{ab}e, c\overline{ab}f,\dots, fe\overline{ab}$. You should also have counted strings like $\overline{ba}cd, \overline{ba}ce, \overline{ba}cf,\dots, fe\overline{ba}$. The phrase "$a$ and $b$ are adjacent" merely means that $ab$ is a substring or that ba is a substring. You only counted where $a$ occurs to the immediate left of $b$ but you should also have counted where $a$ occurs to the immediate right of $b$ as well. – JMoravitz Nov 07 '18 at 18:12
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    Your logic and workings are good but ignored that final step of determining whether $a$ is to the left of $b$ or if $a$ is to the right of $b$. Multiplying the final result by two corrects this. – JMoravitz Nov 07 '18 at 18:13

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