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Home » In how many ways can the letters of the word ‘ARRANGE’ be arranged so that the two R’s are never together?
In how many ways can the letters of the word ‘ARRANGE’ be arranged so that the two R’s are never together?
CBSE | Class 11 | Exercise 16.5 | Maths | Permutations | RD Sharma
In how many ways can the letters of the word ‘ARRANGE’ be arranged so that the two R’s are never together?

There are

    \[7\]

letters in the word

    \[ARRANGE\]

out of which

    \[2\text{ }are\text{ }As,\text{ }2\text{ }are\text{ }Rs\]

and the rest all are distinct.

So by using the formula,

    \[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\]

total number of arrangements

    \[=\text{ }7!\text{ }/\text{ }\left( 2!\text{ }2! \right)\]

    \[=\text{ }\left[ 7\times 6\times 5\times 4\times 3\times 2\times 1 \right]\text{ }/\text{ }\left( 2!\text{ }2! \right)\]

Or,

    \[=\text{ }7\times 6\times 5\times 3\times 2\times 1\]

    \[=\text{ }1260\]

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.

So these

    \[6\]

letters can be arranged in

    \[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\text{ }=\text{ }6!/2!\]

Ways.

The number of words in which all

    \[Rs\]

come together

    \[=\text{ }6!\text{ }/\text{ }2!\]

    \[=\text{ }\left[ 6\times 5\times 4\times 3\times 2! \right]\text{ }/\text{ }2!\]

Or,

    \[=\text{ }6\times 5\times 4\times 3\]

    \[=\text{ }360\]

So, 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

    \[=\text{ }1260\text{ }-\text{ }360\]

    \[=\text{ }900\]

Hence, the total number of arrangements of word

    \[ARRANGE\]

 in such a way that not all R’s come together is

    \[900\]

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