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 ‘k’ lies in the interval.
(a) [1, 5]
(b) (5, 10]
(c) (10, 15]
(d) (15, 20]

Answer 1

Hint: To solve the given question, we will first find out how many distinct letters there in the given word are and how many are the same. Then we will find the total number of 9 lettered words that can be formed without any restriction. Then we will take up two cases – one in which both N are together and others in which both E are together. We will subtract these cases from a total of 9 lettered words formed to get the answer.

Complete step-by-step solution -
The distinct words in the letter ‘MEENANSHU’ are 7 and the same letter pairs are 2 which are E and N. Now, we will find the total number of 9 letter words that can be formed from the word ‘MEENANSHU’ without any restriction. There are a total of 9 letters in the given word and there are two same pairs of letters. Thus,
$$\text{Total Number of Words }={\displaystyle \frac{9!}{2!2!}}$$
Now, we will take up the cases in which either both N are together or both E are together.
Case 1: Both N are together. When both N are together, we will consider NN as a single letter. So the total number of letters is 8. The total number of words formed when both N are together is given by,
$$\text{Total Words in which Ns are together }={\displaystyle \frac{8!}{2!}}$$
There is a 2! In the denominator because there are two Es in the word.
Case 2: Both Es are together. When both Es are together, we will consider EE as a single letter. So, the total number of letters is 8. The total number of words formed when both E are together is given by,
$$\text{Total words in which Es are together }={\displaystyle \frac{8!}{2!}}$$
Now, in the above cases, there will be a situation when both NN and EE are together. So, we will have to subtract that case from the above cases. If both NN and EE are together, we will consider both NN and EE as single letters. Thus, total letters become 7. Thus,
Total words in which Ns and Es are together = 7!
Now, the total number of words obtained from Case 1 and Case 2
$$={\displaystyle \frac{8!}{2!}}+{\displaystyle \frac{8!}{2!}}-7!$$
Now, we will subtract the above numbers from the total to get the required number of words. Thus, we have,
Required number of words = Total – (Words obtained from Case 1 and Case 2)
$$\Rightarrow \text{Required Number of words}={\displaystyle \frac{9!}{2!2!}}-({\displaystyle \frac{8!}{2!}}+{\displaystyle \frac{8!}{2!}}-7!)$$
$$\Rightarrow \text{Required Number of words}={\displaystyle \frac{9!}{2!2!}}-8!+7!$$
Now, it is given in the question that the required number of words is R (7!). Thus, we have,
$$k(7!)={\displaystyle \frac{9!}{2!2!}}-8!+7!$$
$$\Rightarrow ={\displaystyle \frac{9!}{7!2!2!}}-{\displaystyle \frac{8!}{7!}}+{\displaystyle \frac{7!}{7!}}$$
Now, we know that, n! = n (n – 1)! = n (n – 1)(n – 2)!. So, we will get,
$$\Rightarrow k={\displaystyle \frac{9\times 8\times 7!}{7!\times 2\times 2}}-{\displaystyle \frac{8\times 7!}{7!}}+{\displaystyle \frac{7!}{7!}}$$
$$\Rightarrow k={\displaystyle \frac{9\times 8}{2\times 2}}-8+1$$
$$\Rightarrow k=18-8+1$$
$$\Rightarrow k=11$$
Therefore, R belongs to the interval (10, 15].
Hence, the option (c) is the right answer.

Note: It is necessary to subtract the number of ways when both Es and both Ns are together from the total of the cases. This is because there will be a situation when the word formed will be of the form: MANNSEEUH and we have counted this type of words both in Case 1 and Case 2. So, we have subtracted the number of ways obtained when both Ns and both Es are together from the total of the cases.