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Number of permutation of n objects with n1 identical objects of type 1, n2 identical objects of type 2, ......and nk identical objects of type k is $\frac{n!}{n_{1}!n_{2}!....n_{k}!}$

Permutation of letter BANANA are $\frac{6!}{3!2!}= 60$

Option c) is correct

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$\text{Total number of letters in ''BANANA''} = 6$

$\text{Total number of 'A' in the word ''BANANA'' }= 3$

$\text{ Total number of  'N' in the word ''BANANA''}= 2$

$\text{Total distinguishable permutations of the letters ''BANANA'' }=\frac{6!}{3! 2!}=60$
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Distinguishable permutation means the letters won't be repeated. Here BANANA contains 6 letters in which N has been repeated twice and A has been repeated thrice. Hence number of distinguishable permutations = 6!/(3!2!) = 60
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