In how many ways the word arrange be arranged such that the 2 rs come together

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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