## Precalculus (6th Edition) Blitzer

The required solution is $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=\frac{3}{4}$ .
The permutation formula is: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ Applying these formulas into the provided expression, we get: $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=1-\frac{\frac{3!}{\left( 3-2 \right)!}}{\frac{4!}{\left( 4-3 \right)!}}=1-\frac{3!}{1!}\times \frac{1!}{4!}$ $=1-\frac{3\times 2\times 1}{4\times 3\times 2\times 1}$ \begin{align} & =1-\frac{1}{4} \\ & =\frac{3}{4} \\ \end{align}