Question

How many different words using all the letters of ZOONOOZ can be formed if there are no O’s together?

Answer 1

As we know that the above question is of permutations. We can see that in the word ZOONOOZ, there are four O’s , two Z and one N. We have to find the number of ways the word can be arranged so that no O’s are together. Using the concept of permutations we can identify the formula and get the number of ways. The basic formula that can be applied in permutations is $\dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!...}}$.

Complete step by step solution:
We can see that in the word ZOONOOZ there are total $7$ alphabets i.e. $n = 7$. It means the number of possible ways of $11$ alphabets is $11!$.
But since no for O’s can be together, so they can all be taken as one and they can be arranged in $1!$.
Now we have $2Z$ and $1N$, these three letters can be arranged in three ways i.e. $3!$. It gives us value $3 \times 2 \times 1 = 6$ ways.
But here we have two Z’s, they are duplicates, so we need to divide by the number of ways they can arrange themselves which is $2! = 2$.
So the total number of ways it can be arranged is $\dfrac{6}{{2 \times 1!}} = 3$ways.

Hence the total number of required ways so that no O’s are together is $3$.

Note: Note that in these types of problems we first need to simplify the functions. And then we will start finding the angle. Also note that arc is nothing but an inverse function. We chose the sin function domain because in our problem we have the same function.