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