Question

Number of different words that can be formed from the word 'PERMUTATIONS' such that there should be atleast two alphabets between 'M' and 'A' are

• A. $45\times 10!$
  
• B. $54\times 10!$
• C. $90\times 10!$
• D. $11!-10!$

Answer 1

The correct option is A $45\times 10!$


Required number of words = total words - total words having less than two alphabets between 'M' and 'A'
Total words=$\frac{12!}{2!}$
Total words having less than two alphabets between 'M' and 'A=Total words having no alphabets between 'M' and 'A+ Total words having one alphabet between 'M' and 'A

Total words having no alphabets between 'M' and 'A =$\frac{11!}{2!}\times 2!$
Total words having one alphabet between 'M' and 'A $=8\times \frac{10!}{2!}\times 2!\text{when T is not in between}+10!\times 2!\text{when T is in between}$


Total numbers = $\frac{11!}{2!}\times 2!+8\times \frac{10!}{2!}\times 2!+10!\times 2!$
Required number of words=$\frac{12!}{2!}-(\frac{11!}{2!}\times 2!+8\times \frac{10!}{2!}\times 2!+10!\times 2!)$
$=45\times 10!$