Question

The number of $9$ lettered words that can be formed using all the letters of the word 'MEENANSHU' if alike letters are never adjacent, is $k(7!)$. Then ${}^{\prime }{k}^{\prime }$ lies in the interval:

• A. $[1,5]$
• B. $(5,10]$
• C. $(10,15]$
• D. $(15,20]$

Answer 1

The correct option is D $(15,20]$

Total no. of letters in MEENANSHU=9

Total no. of distinct letters in MEENANSHU=7

Total no. of words that can be formed by using all the letters=$\frac{9!}{2!\times 2!}$ [$\because$'E' and 'N' are repeated twice each]

Now, we will calculate total no. of words with alike letters adjacent and for this we consider 2 'N' as one letter and 2 'E' as one other letter

This will reduce total no. of characters to 7 which are M, EE, NN, A, S, H, U

So, total no. of words that can be formed now=$7!$

No, of words with alike letters never adjacent=$\frac{9!}{2!\times 2!}-7!$

$\Rightarrow 7!(\frac{9\times 8}{2\times 2}-1)$

$\Rightarrow 17\times (7!)$

Hence,$l=17\epsilon (15,20]$ which is option $D$