Question

If the letters of the word panchatantra be arranged randomly then what is the probability that there are ten letters betweenr and p

Answer 1

The probability is 0

Answer 2

The required probability is $\dfrac{1}{66}$ .

Step-by-step explanation:

Consider the provided information.

The number of letters in PANCHATANTRA is 12.

A is repeated 4 times, N for 2 times and T for 2 times.

The total number of ways are: $n(s)=\dfrac{12!}{2!2!4!}$

We need to find the probability that there are ten letters between R and P.

We have 2 ways either letters start with R and end with P or it can be start from P and end with R.

The number of ways to arrange rest of 10 numbers are: $n(e)=\dfrac{10!}{2!2!4!}$

Therefore, the required probability is:

$\dfrac{n(e)}{n(s)}=\dfrac{\dfrac{10!}{2!2!4!}}{\frac{12!}{2!2!4!}}\times 2$

$\dfrac{n(e)}{n(s)}=\dfrac{10!}{12!}\times2$

$\dfrac{n(e)}{n(s)}=\dfrac{2}{11\times12}$

$\dfrac{n(e)}{n(s)}=\dfrac{1}{66}$

Hence, the required probability is  $\dfrac{1}{66}$ .

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