How many ways we can arrange a letters of the word ENGINEERING so that no two Es are come together?

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Text Solution

Solution : there are 3 E,3N,2G,2I and 1R.<br> Total no. of arrangements when all three Ns come together=`(9!)/(3!2!2!)`....(1)<br> Total no. of arrangements when all three Ns and Es come together=`(7!)/(2!2!)`....(2)<br> Required ways=>(1)-(2)=`13860`

The number of ways the N's can come together The letters of ENGINEERING arranged in alphabetical order is E,E,E,G,G,I,I,N,N,N,R The number of ways the three N's can come together is the number of distinguishable permutations of these 9 things, where the (NNN) is considered as a single "thing". E,E,E,G,G,I,I,(NNN),R Since there are 3 indistinguishable E's, 2 indistinguishable G's, and 2 indistinguishable I's, the number is:

From this 15120 we must subtract the number of ways 2 or 3 E's can come together. First we will calculate the number of distinguishable arrangements of these 7 things: (EE),G,G,I,I,(NNN),R where one E is missing, and then we'll calculate how many ways we can insert the third E into each one of those. Here the (EE) and the (NNN) are each considered as just one "thing". Since there are 2 indistinguishable G's, and 2 indistinguishable I's, the number is:

= 1260 An example would be GG(EE)II(NNN)R. Let's put a space before and after each letter or "thing" to indicate feasible places to insert the third E. We'll number the spaces below each space, like this: _G_G_(EE)_I_R_(NNN)_I_ 1 2 3 4 5 6 7 8 We might think at first that we could insert the 3rd E into any of the 8 spaces. However we would not be able to distinguish between placing the third E in positions 3 or 4 above. That is, we could not tell the difference between the result of inserting the third E just before, or just after, the (EE). So there are 1 less than 8, or only 7 places that we can insert the third E and have a distinguishable permutation. So we multiply 1260 by the 7 ways to insert the third E. So 1260�7 = 8820. That is the number which we must subtract from the 15120. Final answer = 15120 - 8820 = 6300 distinguishable permutations. Edwin

Answer

Verified

Hint: 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.