Answer
The required solution is $1-\frac{{}_{5}{{P}_{3}}}{{}_{10}{{P}_{4}}}=\frac{83}{84}$ .
Work Step by Step
The Permutation formula is:
${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$
Applying this formulas into the provided expression, we get:
$1-\frac{{}_{5}{{P}_{3}}}{{}_{10}{{P}_{4}}}=1-\frac{\frac{5!}{\left( 5-3 \right)!}}{\frac{10!}{\left( 10-4 \right)!}}=1-\frac{5!}{2!}\times \frac{6!}{10!}$
$=1-\frac{5\times 4\times 3\times 2\times 1}{2\times 1}\times \frac{6\times 5\times 4\times 3\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}$
$\begin{align}
& =1-\frac{60}{5040} \\
& =\frac{83}{84}
\end{align}$
$1-\frac{{}_{5}{{P}_{3}}}{{}_{10}{{P}_{4}}}=\frac{83}{84}$
