Counting

$5 \times 4 \times 3 \times 2 \times 1 = 120$. This is the number of ways of arranging 5 distinct objects.

All five letters are distinct, so the answer is $5! = 120$.

$\dfrac{7 \times 6}{2 \times 1} = 21$. Order does not matter in a combination.

Circular arrangements give $(n-1)! = 5! = 120$, since only relative positions matter.

$\binom62 \times \binom52 = 15 \times 10 = 150$. Choose the men and the women independently, then multiply.

$P(A)P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B)$, so the events are independent. Equivalently $P(A|B) = 0.2/0.5 = 0.4 = P(A)$.