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The no. of arrangements of the letters of the word 'BANANA' in which the 2 $N^s$ do not come adjacently is?

$\begin{array}{1 1} 40 \\ 60 \\ 80 \\ 100 \end{array}$

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Total no. of arrangements of the letters of the word 'BANANA'=$\large\frac{6!}{3!.2!}$
$=60\:arrangements$
Considering 2 $N^s$ together the no. of arrangements = $\large\frac{5!}{3!}$
$=20\:arrangements$
$\therefore$ The no. of arrangements in which both the $N^s$ do not come adjacently
$=60-20=40\:arrangements.$
answered Aug 12, 2013 by rvidyagovindarajan_1
edited Aug 4, 2014 by sharmaaparna1
 

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