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Home » How many words can be formed with the letters of the word ‘PARALLEL’ so that all L’s do not come together?
How many words can be formed with the letters of the word ‘PARALLEL’ so that all L’s do not come together?
CBSE | Class 11 | Exercise 16.5 | Maths | Permutations | RD Sharma
How many words can be formed with the letters of the word ‘PARALLEL’ so that all L’s do not come together?

Given:

The word

    \[PARALLEL\]

There are

    \[8\]

letters in the word

    \[PARALLEL\]

out of which

    \[2\text{ }are\text{ }As,\text{ }3\text{ }are\text{ }Ls\]

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

    \[=\text{ }8!\text{ }/\text{ }\left( 2!\text{ }3! \right)\]

    \[=\text{ }\left[ 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right]\]

    \[/\text{ }\left( 2\times 1\times 3\times 2\times 1 \right)\]

Or,

    \[=\text{ }8\times 7\times 5\times 4\times 3\times 1\]

    \[=\text{ }3360\]

Now, let us consider all

    \[Ls\]

together as one letter, so we have

    \[6\]

letters out of which

    \[A\]

repeats

    \[2\]

times and others are distinct.

These

    \[6\]

letters can be arranged in

    \[6!\text{ }/\text{ }2!\]

Ways.

The number of words in which all L’s come together

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

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

Or,

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

    \[=\text{ }360\]

So, now the number of words in which all

    \[Ls\]

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

    \[=\text{ }3360\text{ }-\text{ }360\text{ }=\text{ }3000\]

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