College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.6 - Counting Theory - 7.6 Exercises - Page 679: 26

Answer

$75,287,520$

Work Step by Step

Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ the given expression, $ _{100}C_{5} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{100!}{5!(100-5)!} \\\\= \dfrac{100!}{5!95!} \\\\= \dfrac{100(99)(98)(97)(96)(95!)}{5(4)(3)(2)(1)(95!)} \\\\= \dfrac{\cancel{100}^5(\cancel{99}^{33})(\cancel{98}^{49})(97)(96)(\cancel{95!})}{\cancel{5(4)}(\cancel{3})(\cancel{2})(1)(\cancel{95!})} \\\\= \dfrac{75287520}{1} \\\\= 75,287,520 \end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.