Question

In the word $'ENGINEERING'$ if all $E's$ are not together and $N's$ are always together, then number of permutations is
$A. = \dfrac{{9!}}{{2!2!}} - \dfrac{{7!}}{{2!2!}}$
$B. = \dfrac{{9!}}{{3!2!}} - \dfrac{{7!}}{{2!2!}}$
$C. = \dfrac{{9!}}{{3!2!2!}} - \dfrac{{7!}}{{2!2!2!}}$
$$D. = \dfrac{{9!}}{{3!2!2!}} - \dfrac{{7!}}{{2!2!}}$$$$$$

Answer 1

In this question, we have to find out the total number of ways in which no $E's$ will come together and $N's$ are always together. To solve the question, note down the letters in a row and work out what letters are repeated and how many times. Take out no. of ways in which all $N's$ are together. Take all N as one unit and work out the total no. of wages. Repeat this way to solve out no. of ways in which $E's$ will together. Then subtract the letter from the former to take out the ways in which no $E's$ come together.

Complete step-by-step answer:
Writing down the letters of $ENGINEERING$-
$E,E,E \\\ N,N,N \\\ G,G \\\ I,I \\\ R \\\$
$E$ comes $3$ times, $N$ comes $3$ times, $G$ comes $2$ times, $I$ comes $2$ times & $R$ comes $1$ time.
Now, when $N's$ come together.
Take all $N$ as one, then total numbers of letters become $9$ and they arrange $9!$ ways.
But some repeated letters are present.
So, no. of ways arrangement needs when $N$ remarks together $= \dfrac{{9!}}{{3!2!2!}}$.
When $N's$ & $E's$ are come together $= \dfrac{{7!}}{{2! \times 2!}}$.
$\therefore$ Required number of arrangements are $= \left( {\dfrac{{9!}}{{3!2!2!}} - \dfrac{{7!}}{{2! \times 2!}}} \right)$.

So, the correct answer is “Option D”.

Note: The question asked was from the topic permutation & combination. When we are arranging some object in different ways, they are permuted. While if we select some from many given objects are called combinations. This is totally a conceptual based question.