**Question **: There are only two women among 20 persons taking part in a pleasure trip. The 20 persons are divided into two group, each group consisting of 10 persons. Then the probability that the two women will be in the same group is

(A) 9/19

(B) 9/38

(C) 9/35

(D) none of these

**Answer **: option (A)

**Solution **:

The number of ways in which 20 peoples can be divided into two equal groups is

n\left( s\right) =\dfrac{20!}{10!10!2!}
The number of ways in which 18 peoples can be divided into groups of 10 and 8 is

n\left( A\right) =\dfrac{18!}{10!8!}
therefore, P\left( E\right) =\dfrac{18!}{10!8!}\times \dfrac{10!10!2!}{20!}=\dfrac{10\times 9\times 2}{20\times 19}=\dfrac{9}{19}.

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