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How many words can be formed by re-arranging the letters of the word “PROBLEMS” such that $P$ and $S$ occupy the first and last position respectively? (Note: The words thus formed need not be meaningful)
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'PROBLEMS' is an eight-letter word where none of the letters repeat.

We know that “$r$” objects, where all are distinct, can be reordered in $r!$ ways.

Therefore, if there had been no constraints, the letters of 'PROBLEMS' can be reordered in $8!$ ways.

However, the question states that '$P$' and '$S$' should occupy the first and last position respectively.

Therefore, the first and last position can be filled in only one way.

The remaining $6$ positions in between $P$ and $S$ can be filled with the $6$ letters in $6!$ ways.

Rearrangement Concepts: Number of ways to reorder objects:

  1. $r$ distinct objects can be reordered in $r!$ Ways.
  2. $r$ similar objects can be reordered in only $1$ way.
  3. 3 $r$ objects, of which $x$ are alike, can be reordered in $r!/x!$ ways.
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“PROBLEMS” word has 8 distinct word ‘P’,’R’,’O’,’B’,’L’,’E’,’M’,’S’ .

Now we need to re-arrange the word where first letter is ‘P’ and last letter is ‘S’ .

P_ _ _ _ _ _S   , so we have 6 letter left which can be rearrange in $6!$ ways.

So , Total $720$ words are possible .
Answer:

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