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

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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

$=60-20=40\:arrangements.$

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