Yes. This is what I have written. put the N value in N*logN to get the ans.

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The given number N can be represented in logN bits which is n. (given) { log N = n or N = $2^{n}$}

So, to represent N!, we need at least logN! bits.

And, according to Stirling’s approximation log n! $\approx$ n log n

We can write logN! = N log N

**Therefore, we can write $2^{n} n$ are the bits from N! numbers can be represented.**

In the options, none of them is matching.

But if you still want the correct option, you can go with the most suitable answer matched that is $2^{n}$.