Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 11 - Probability and Statistics - 11-1 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 678: 19

Answer

$3,003$

Work Step by Step

Using $n!=n(n-1)(n-2)\cdot...(3)(2)(1),$ the given expression, $ \dfrac{15!}{10!\text{ }5!} ,$ simplifies to \begin{align*}\require{cancel} & \dfrac{15(14)(13)(12)(11)(10!)}{10!\text{ }5(4)(3)(2)(1)} \\\\&= \dfrac{15(14)(13)(12)(11)(\cancel{10!})}{\cancel{10!}\text{ }5(4)(3)(2)(1)} \\\\&= \dfrac{15(14)(13)(\cancel{12})(11)}{5(\cancel4)(\cancel3)(2)(1)} \\\\&= \dfrac{\cancel{15}^3(14)(13)(11)}{\cancel5(2)(1)} \\\\&= \dfrac{3(\cancel{14}^7)(13)(11)}{(\cancel2)(1)} \\\\&= 3(7)(13)(11) \\&= 3003 .\end{align*} Hence, the given expression evaluates to $ 3,003 .$
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