College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 52

Answer

Yes, the statement is true.

Work Step by Step

$\displaystyle \left(\begin{array}{l}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}$ $\displaystyle \left(\begin{array}{l}n\\n-r \end{array}\right)=\frac{n!}{(n-r)!r!}$ Thus we have shown that (for $0\leq r\leq n$): $\displaystyle \left(\begin{array}{l}n\\r\end{array}\right)= \left(\begin{array}{l}n\\n-r \end{array}\right)$
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