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Find the number of combinations of $3$ items from a set of $5$.
Hint: ${}^n \mathrm{C}_r = \dfrac{n!}{r! \, (n-r)!}$
${}^{5} \mathrm{C}_{3} = \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}!}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{54px}}~}! \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{54px}}~}!}$ ${}^{5} \mathrm{C}_{3} = \dfrac{5!}{3! \times 2!}$
Cancel and calculate. Hint: Multiply the numbers down from $n$ in the numerator, and down from $r$ in the denominator.
${}^{5} \mathrm{C}_{3} = \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{35px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{35px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{35px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{35px}}~}} = \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{80px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{80px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{DR}}\hspace{80px}}~}$ ${}^{5} \mathrm{C}_{3} = \dfrac{5 \times 4}{2 \times 1} = \dfrac{20}{2} = 10$