Answer
$1,n$
Work Step by Step
For $n\choose j$ the correct formula is $\frac{n!}{(n-j)!j!}$.
Hence ${n\choose 0}=\frac{n!}{(n-0)!0!}=\frac{n!}{(n)!\cdot1}=1$ , ${n\choose 1}=\frac{n!}{(n-1)!1!}=\frac{n!}{(n-1)!\cdot1}=n$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.5 The Binomial Theorem - 12.5 Assess Your Understanding - Page 836: 2
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