Solveeit Logo
Login

Question

Question: Find the number of ways the letters of the word ‘MUNMUN’ can be arranged so that no two alike letter...

Find the number of ways the letters of the word ‘MUNMUN’ can be arranged so that no two alike letters are together.

Explanation

Solution

We first find the number of ways the word ‘MUNMUN’ can be arranged without any condition. We find the conditional events where we find the same words being together using the inclusion and exclusion theorem. We subtract them to find the final solution.

Complete step by step answer:
We first find the number of ways the word ‘MUNMUN’ can be arranged without any condition.The number of ways will be 6!2!×2!×2!=7208=90\dfrac{6!}{2!\times 2!\times 2!}=\dfrac{720}{8}=90.Now we try to find the number of ways the word ‘MUNMUN’ can be arranged so that no two alike letters are together.We take the events of 2 ‘M’s, 2 ‘U’s, 2 ‘N’s as events P, Q, R respectively.n(P∪Q∪R)n\left( P\cup Q\cup R \right) denotes the event of two similar letters coming together.

We use inclusion and exclusion theorem which tells us that
n(P∪Q∪R)=n(P)+n(Q)+n(R)−n(P∩R)−n(P∩Q)−n(Q∩R)+n(P∩Q∩R)n\left( P\cup Q\cup R \right)=n\left( P \right)+n\left( Q \right)+n\left( R \right)-n\left( P\cap R \right)-n\left( P\cap Q \right)-n\left( Q\cap R \right)+n\left( P\cap Q\cap R \right).
Now we find the individual values.The number of events for n(P),n(Q),n(R)n\left( P \right),n\left( Q \right),n\left( R \right) are all equal. This denotes the event where the individual words come together.So,
n(P)=n(Q) ⇒n(P)=n(R) ⇒n(P)=5!2!×2! ⇒n(P)=30n\left( P \right)=n\left( Q \right) \\\ \Rightarrow n\left( P \right) =n\left( R \right) \\\ \Rightarrow n\left( P \right) =\dfrac{5!}{2!\times 2!} \\\ \Rightarrow n\left( P \right) =30

The number of events for n(P∩R),n(P∩Q),n(Q∩R)n\left( P\cap R \right),n\left( P\cap Q \right),n\left( Q\cap R \right) are all equal. This denotes the event where two individual words come together.So,
n(P∩R)=n(P∩Q) ⇒n(P∩R)=n(Q∩R) ⇒n(P∩R)=4!2! ⇒n(P∩R)=12n\left( P\cap R \right)=n\left( P\cap Q \right) \\\ \Rightarrow n\left( P\cap R \right) =n\left( Q\cap R \right) \\\ \Rightarrow n\left( P\cap R \right) =\dfrac{4!}{2!} \\\ \Rightarrow n\left( P\cap R \right) =12
Lastly, we get n(P∩Q∩R)=3!=6n\left( P\cap Q\cap R \right)=3!=6 denotes the event of all three alike letters coming together.The final outcome is
n(P∪Q∪R)=3×30−3×12+6 ∴n(P∪Q∪R)=60n\left( P\cup Q\cup R \right)=3\times 30-3\times 12+6 \\\ \therefore n\left( P\cup Q\cup R \right) = 60
We subtract and get 90−60=3090-60=30 as the number of ways the word ‘MUNMUN’ can be arranged so that no two alike letters are together.

Hence, in 30 different ways the word ‘MUNMUN’ can be arranged so that no two alike letters are together.

Note: We need to remember that the events n(P),n(Q),n(R)n\left( P \right),n\left( Q \right),n\left( R \right) are the total events where the primary condition is that one set of alike words come together, the rest of the condition may or may not happen.