The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.

Tricky one.

5 sides of that pentagon are \(a,b,c,d,e\)

The totale numbers of possibles pentagons are \(5 \times 4 \times 3 \times 2 \times 1 =5!\)

Note that each possible selection of \((a,b,c,d,e)\), there are 4 other possible selections that are formed by arranged around each element. For example these 5 selections are considered the same pentagons
\(\begin{split}
2,3,4,5,6\\
6,2,3,4,5\\
5,6,2,3,4\\
4,5,6,2,3\\
3,4,5,6,2\\
\end{split}\)

Hence, in order to count the possible different selections, we need to divide the result by 5: \(5!/5=4!\)

Note that each pentagon could be fliped over, so both are considered the same. For example, these 2 selections are considered the same pentagons
\(\begin{split}
2,3,4,5,6\\
6,5,4,3,2\\
\end{split}\)

So, we also need to divide the result by 2 to avoid flipped over ones: \(4!/2=12\)

The final result are 12. The answer is B.

If the result are 24, this means that it covers the flipped over pentagons.