Given: 5 objects taken 2 at a time
[tex]\begin{gathered} \text{Permutation} \\ _5P_2=\frac{n!}{(n-r)!}=\frac{5!}{(5-2)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3!}=\frac{120}{3\cdot2\cdot1}=\frac{120}{6}=20 \\ \\ \\ \text{Combination} \\ _5C_2=\frac{n!}{r!(n-r)!}=\frac{5!}{2!(5-2)!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2!\cdot3!}=\frac{120}{(2\cdot1)(3\cdot2\cdot1)}=\frac{120}{12}=10 \end{gathered}[/tex]
