Question

If 100 boys are arranged at random along a circle then the odds against to arrange two specific boys of those 100 come together is
A. 2:97
B. 97:2
C. 2:98
D. 98:2

Answer 1

First find the total number of ways in which boys can be arranged in a circle. Then, find the probability of arranging two specific boys together. Next, find the probability of not arranging two specific boys together. At, determine the ratio of arranging boys such that two specific boys do not come together to two specific boys coming together.

Complete step-by-step answer:
We have to arrange 100 boys in a circle.
It is known that if there are $n$ objects that need to be arranged in a circle, there are $\left( {n - 1} \right)!$ ways of doing it.
Hence, we can arrange 100 boys in a circle in $\left( {100 - 1} \right)! = 99!$ ways.
We have to find the number of ways of arranging 100 boys when two specific boys come together.
Let two specific boys be one group.
Now, we have to arrange 98 remaining boys and 1 group.
That is , we have to arrange 99 objects in a circle.
$\left( {99 - 1} \right)! = 98!$switch their positions.
Therefore, total number of ways in which 2 specific boys can be arranged together is $2 \times 98!$
Also, the boys can
We will find the probability of arranging two specific boys together.
And probability of an event is equal to $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{total possible outcomes}}}}$
That is $P\left( A \right) = \dfrac{{2 \times 98!}}{{99!}}$
Now, we know that $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).......3.2.1$
$\dfrac{{2 \times 98!}}{{99.98!}} = \dfrac{2}{{99}}$
We will also have to find the probability of not arranging two specific boys together.
And we know that $P\left( A \right) + P\left( {\overline A } \right) = 1$
$\dfrac{2}{{99}} + P\left( {\overline A } \right) = 1 \\\ \Rightarrow P\left( {\overline A } \right) = 1 - \dfrac{2}{{99}} \\\ \Rightarrow P\left( {\overline A } \right) = \dfrac{{97}}{{99}} \\\$
Next, find the ratio of arranging boys such that two specific boys do not come together to two specific boys.
$\dfrac{{97}}{{99}}:\dfrac{2}{{99}} = 97:2$
Hence, option B is correct.

Note: If $n$ objects are to be arranged in $n$ places, then there are $n!$ ways, but if there are $n$ ways that needs to be arranged in a circle, then there are $\left( {n - 1} \right)!$ ways. Also, the sum of probability of an event and probability of not happening an event is equal to 1.