Question

The letters of the word PROBABILITY are rearranged in a random order. What is the probability that the

letters P and L have exactly 3 letters between them?

Answer 1

The word PROBABILITY has 11 letters and 2 repeating letters.

1 P, 1 R, 1 O, 2 B, 1 A, 2 I, 1 L, 1 T and 1 Y.

Total no. of meaningful and non-meaningful words that can be obtained by rearranging these letters

$= \frac{11!}{2! \times 2!}$

Now we have to find the no. of words in which the letters P and L have exactly 3 letters in between them. Given below are the chances for that:

PxxxLxxxxxx

xPxxxLxxxxx

xxPxxxLxxxx

xxxPxxxLxxx

xxxxPxxxLxx

xxxxxPxxxLx

xxxxxxPxxxL

There are 7 chances.

Except P and L, there are 9 letters with 2 repeating letters:

1 R, 1 O, 2 B, 1 A, 2 I, 1 T and 1 Y.

No. of chances (7) multiplied by no. of meaningful and non-meaningful words that can be obtained by these 9 letters gives the no. of words in which P and L have exactly 3 letters in between them, i. e.,

$\frac{9!}{2! \times 2!} \times 7 \\ \\ = \frac{9! \times 7}{2! \times 2!}$

Probability

$= \frac{9! \times 7}{2! \times 2!} \div \frac{11!}{2! \times 2!} \\ \\ = \frac{9! \times 7}{2! \times 2!} \times \frac{2! \times 2!}{11!} \\ \\ = \frac{9! \times 7}{11!} \\ \\ = \frac{9! \times 7}{11 \times 10 \times 9!} \\ \\ = \frac{7}{11 \times 10} \\ \\ = \frac{7}{110}$

So the probability is 7/110. I think it is the right answer.

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