The letters of word ARRANGE are arranged such that two Rs are never together

Complete step by step answer:
Given:
The word ARRANGE. It has a total of 7 words out of which there are two A’s and two R’s and the rest three words are different.
Now, when we have $n$ objects such that $p$ items are of one type and $q$ are of another type and the rest $r$ are different objects, so $n=p+q+r$ . Then, we can arrange them in $\dfrac{n!}{\left( p! \right)\left( q! \right)}$ ways.
Now, in our question, we have word ARRANGE where total 7 words are there out of which 2 are A’s and 2 are R’s and rest 3 are different then, the total number of ways in which letters of the given word can be arranged will be $\dfrac{7!}{\left( 2! \right)\left( 2! \right)}=1260$ .
(a) The two R’s are never together:
Now, for this case we will club the two R’s together and treat it as one single letter then we will have a total 6 words out of which 2 are A’s and rest 4 are of different types. And we can arrange them in $\dfrac{6!}{\left( 2! \right)}=360$ the number of ways. Then, when we subtract this from the total number of arrangements that is 1260 then we will get the number of arrangements in which the two R’s are never together.
Thus, we can arrange the letter of the word ARRANGE such that two R’s are never together is equal to $1260-360=900$ .
Hence, 900 such arrangements are possible in which two R’s are never together.
(b) The two A’s are together but not two R’s:
Now, for this case we will club the two A’s together and treat it as one single letter then we will have a total 6 words out of which 2 are R’s and the rest 4 are of different types. And we can arrange them in $\dfrac{6!}{\left( 2! \right)}=360$ the number of ways. Then, the number of ways in which 2 A’s are together is 360.
Now, club the two A’s and treat it as one single letter and the two R’s and treat it as one single letter then we will have a total 5 different words which can be arranged in $5!=120$ ways. Then, the number of ways in which two A’s and two R’s both are together is 120.
Now, when we subtract the number of ways in which two A’s and two R’s both are together from the number of ways in which 2 A’s are together then we will get the number of ways in which the two A’s are together but not two R’s and that is $360-120=240$ ways.
Hence, 240 such arrangements are possible in which two A’s are together but not two R’s.
(c) Neither two A’s nor the two R’s are together.
Now, when we subtract the number of ways in which the two A’s are together but not two R’s from the number of arrangements in which the two R’s are never together then we will get the number of arrangements in which neither two A’s nor the two R’s are together and that is $900-240=660$ ways.
Hence, 660 such arrangements are possible in which neither two A’s nor the two R’s are together.

Note: Here, the student must take care while making different cases that are possible and not directly apply the formula for the total number of different arrangements of a certain number of different objects in a linear arrangement.

Here's 7 letters in the word ‘ARRANGE’ out of which 2 are A’s, 2 are R’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 7! / (2! 2!)

= [7 × 6 × 5 × 4 × 3 × 2 × 1] / (2! 2!)

= 7 × 6 × 5 × 3 × 2 × 1

= 1260

Now, let us consider all R’s together as one letter, there are 6 letters remaining. Out of which 2 times A repeats and others are distinct.

Therefore these 6 letters can be arranged in n!/ (p! × q! × r!) = 6!/2! Ways.

Number of words in which all R’s come together = 6! / 2!

= [6 × 5 × 4 × 3 × 2!] / 2!

= 6 × 5 × 4 × 3

= 360

Therefore, now the number of words in which all L’s do not come together = total number of arrangements – The number of words in which all L’s come together

= 1260 – 360

= 900

Thus, the total number of arrangements of word ARRANGE in such a way that not all R’s come together is 900.

Solution : The letters of word ARRANGE can be rewritten as
A R N G E
A R
So we have 2 A's and 2 R's , and total 7 letters.
(i) Total number of words is `(7!)/(2!2!)=1260`.
The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g.,
The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g.,
(R R),A,A,N,G,E
Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is
`(7!)/(2!2!)-(6!)/(2!)=900`
(ii) The number of words in which both A's are together is `6!//2!=360`, e.g.,
(A A),R,R,N,G,E
The number of words in which both A's and both R's are together is 5!=120, e.g.,
(A A), (R R), N,G,E
Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240.
(iii) There are in all 900 words in each of which the two R's are never together. Consider any such word. Either the two A's are together or the two A's not together. But the number of all such arrangements in which the two A's are together is 240. Hence, the number of all such arrangements in which the two A's not together is 900-240=660.

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