#### Answer

9

#### Work Step by Step

We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_9C_1 = \frac{9!}{(9-1)!1!}$
$\frac{ 9!}{8!}$
We also know that:
$x! = x(x-1)(x-2)...(1)$
Thus, we have:
9

Published by
Prentice Hall

ISBN 10:
0133281140

ISBN 13:
978-0-13328-114-9

9

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