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Chapter 11: Problem 25

Evaluate each expression. \(\frac{{ }_{7} C_{3}}{{ }_{5} C_{4}}-\frac{98 !}{96 !}\)

Short Answer

Expert verified
-9499

Step by step solution

01

Evaluate \(_{7} C_{3}\)

Replace \(_{7} C_{3}\) with the formula as \(\frac{7!}{3!(7-3)!}\). Now it results in \(\frac{7!}{3!4!}\). It further simplifies to \(\frac{7*6*5*4!}{3*2*1*4!}\). Cancel out the \(4!\) terms for the final answer of 35.
02

Evaluate \(_{5} C_{4}\)

Substitute \(_{5} C_{4}\) with the formula as \(\frac{5!}{4!(5-4)!}\). Now it results in \(\frac{5!}{4!1!}\). On simplification, it gives \(\frac{5*4!}{4!*1}\). The \(4!\) terms cancel to give the result as 5.
03

Evaluate \(\frac{98!}{96!}\)

Solve \(\frac{98!}{96!}\) as \(\frac{98*97*96!}{96!}\). Here, the \(96!\) terms cancel out resulting in 98*97 which equals 9506.
04

Substitute and subtract the two results

Substitute results from step 1 and step 2 back into the first part of expression, then subtract the result from the second part of expression. \(\frac{35}{5}-9506 = 7-9506 = -9499\)

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