Question

1. evaluate \\(\frac{_8c_4}{_8c_5}\\). \\(\frac{_8c_4}{_8c_5} = \\) enter your next step here

Explanation:

Step1: Recall the combination formula

The formula for combinations is \( _nC_r=\frac{n!}{r!(n - r)!} \). So we can write \( _8C_4=\frac{8!}{4!(8 - 4)!} \) and \( _8C_5=\frac{8!}{5!(8 - 5)!} \).

Step2: Simplify the ratio

\(\frac{_8C_4}{_8C_5}=\frac{\frac{8!}{4!4!}}{\frac{8!}{5!3!}}\). The \(8!\) terms will cancel out, so we have \(\frac{5!3!}{4!4!}\).

Step3: Expand the factorials

We know that \(n!=n\times(n - 1)\times\cdots\times1\), so \(5!=5\times4!\) and \(4!=4\times3!\). Substituting these in, we get \(\frac{5\times4!\times3!}{4!\times4\times3!}\).

Step4: Cancel out common terms

The \(4!\) and \(3!\) terms cancel out, leaving us with \(\frac{5}{4}\).

Answer:

\(\frac{5}{4}\)