Answer
always
Work Step by Step
We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_nC_1 = \frac{n!}{(n-1)!1!}$
$\frac{ n*(n-1)!}{(n-1)!}= n$
So, it is always true
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 768: 62
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