Answer
The required solution is $\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{21}{44}$ .
Work Step by Step
The combination formula ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$.
Applying this formulas into the provided expression, we get:
$\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{\frac{5!}{1!\left( 5-1 \right)!}\times \frac{7!}{2!\left( 7-2 \right)!}}{\frac{12!}{3!\left( 12-3 \right)!}}$
$=\frac{5!\times 7!}{4!\times 2!\times 5!}\times \frac{3!\times 9!}{12!}$
$=\frac{7\times 6\times 5\times 4!}{4!\times 2}\times \frac{9!\times 3\times 2\times 1}{12\times 11\times 10\times 9!}$
$=\frac{21}{44}$
$\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{21}{44}$
