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Class11
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Permutations and Combinations
The smallest positive natural number $n$ such that $n!$ is divisible by $990$ is ?
$\begin{array}{1 1} 9 \\ 11 \\ 33 \\ 99 \end{array}$
jeemain
math
class11
ch7
permutations-and-combinations
fundamental-principle-of-counting
easy
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asked
Aug 8, 2013
by
rvidyagovindarajan_1
edited
Apr 5, 2016
by
meena.p
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1 Answer
$990=9\times 10\times 11$
$\therefore $ the least natural number so that $n!$ is divisible by $990$ is 11
answered
Aug 8, 2013
by
rvidyagovindarajan_1
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