In what ways the letters of the word “PUZZLE” can be arranged to form the different new words so that the vowels always come together?

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?

• 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. – 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 $$2!$$.