Q.Evaluate $^{6}P_{2}$.

$^{n}P_{r}$ counts ORDERED arrangements of $r$ objects out of $n$, computed as the top $r$ descending factors of $n!$.

$^{6}P_{2} = \dfrac{6!}{(6-2)!} = \dfrac{6!}{4!} = 6 \times 5 = 30$.

Apply the permutation formula with $n=6$, $r=2$:

Write $6! = 6\times5\times4!$, so the $4!$ cancels:

Check by direct counting (first-principles method): to arrange 2 out of 6 distinct objects into 2 ordered positions, the 1st position can be filled in 6 ways and the 2nd position (one object now used) in 5 ways: $6 \times 5 = 30$. Both methods agree.

Direct counting: the 1st of the two ordered positions can be filled in $6$ ways, the 2nd in the remaining $5$ ways, giving $6\times5=30$ directly — the same descending-factors idea the formula is built from, without writing out $6!/4!$ explicitly.

Confusing $^{n}P_{r}$ with $^{n}C_{r}$ and dividing by $r!$ when the problem calls for an ORDERED count (e.g. answering $15$ instead of $30$ here by mistakenly using the combination formula). Also, computing $(n-r)!$ as $(6-2)! = 4! = 24$ correctly but then forgetting to actually divide $6!$ by it, or dividing $6!=720$ incorrectly.