#### Answer

1365

#### Work Step by Step

We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_{15}C_4 = \frac{15!}{(15-4)!4!}$
$\frac{ 15!}{4!11!}$
$\frac{15*14*13*12*11!}{4!11!}$
$\frac{15*14*13*12}{4*3*2*1}$
1365

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Published by
Prentice Hall

ISBN 10:
0133281140

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

1365

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