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?

  • 1
    $\begingroup$ 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

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 $2!$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.