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?
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