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I tried this way: Taking 'UE' together and remaining 'PZZL'. So, as 'Z' is repeating, I divide it by $2!$. My answer: $(5!/2!)\cdot 2! = 120$. Is it right? And why do we divide by its factorial if some letters are repeating?

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    $\begingroup$ We divide by $k!$ if a letter appears $k$ times because the factorial at the numerator assumes that all letters are distinguishable, which they are not. $\endgroup$ – Fabio Somenzi Aug 2 at 5:53
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Take 'UE' together and 'PZZLE' is remaining

The $E$ should be together with $UE$, so in fact you have something like $(UE)PZZL$ or $(EU)PZZL$.

So, as 'Z' is repeating I divide it by 2!

That's correct. If we call the two letters $Z_1$ and $Z_2$, there are $2!$ ways to arrange the $Z$'s, but there is only one distinguishable combination: you cannot tell $Z_1$ and $Z_2$ apart.

My answer :- (5!/2!)*2! = 120. Is it right?

Yes. There are $5!$ ways to arrange the $5$ objects $UE$, $P$, $Z$, $Z$ and $L$. $U$ and $E$ are interchangeable so we can multiply by $2!$, but as earlier said the two $Z$'s are not, so we divide by $2!$.

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