Answer
$\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{3}{68}$.
Work Step by Step
The combination formula is:
${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
Applying this formulas into the provided expression, we get:
$\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{\frac{4!}{2!\left( 4-2 \right)!}\times \frac{6!}{1!\left( 6-1 \right)!}}{\frac{18!}{3!\left( 18-3 \right)!}}$
$=\frac{4!\times 6!}{2!\times 2!\times 5!}\times \frac{3!\times 15!}{18!}$
$=\frac{4\times 3\times 2\times 6\times 5!}{5!\times 2\times 2}\times \frac{15!\times 3\times 2\times 1}{18\times 17\times 16\times 15!}$
$=\frac{3}{68}$
$\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{3}{68}$
