Question

evaluate the expression.

\\( \frac { _ { 4 } c _ { 2 } \cdot _ { 5 } c _ { 1 } } { _ { 15 } c _ { 3 } } \\)

\\( \frac { _ { 4 } c _ { 2 } \cdot _ { 5 } c _ { 1 } } { _ { 15 } c _ { 3 } } = \square \\) (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate combination values

Use the combination formula \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\).
For \(_{4}C_{2}\):
\(_{4}C_{2}=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3\times2!}{2\times1\times2!}=6\)
For \(_{5}C_{1}\):
\(_{5}C_{1}=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=\frac{5\times4!}{1\times4!}=5\)
For \(_{15}C_{3}\):
\(_{15}C_{3}=\frac{15!}{3!(15 - 3)!}=\frac{15!}{3!12!}=\frac{15\times14\times13\times12!}{3\times2\times1\times12!}=455\)

Step2: Substitute values into the expression

\(\frac{_{4}C_{2}\cdot_{5}C_{1}}{_{15}C_{3}}=\frac{6\times5}{455}\)
\(=\frac{30}{455}\)
Simplify the fraction by dividing numerator and denominator by 5:
\(=\frac{6}{91}\)

Answer:

\(\frac{6}{91}\)