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Question: 2n boys are randomly divided into two subgroups containing n boys each .The probability that the two...

2n boys are randomly divided into two subgroups containing n boys each .The probability that the two tallest boys are in different groups is
A) n2n−1\dfrac{{\text{n}}}{{2{\text{n}} - 1}}
b) n−12n−1\dfrac{{{\text{n}} - 1}}{{2{\text{n}} - 1}}
C) 2n−14n2\dfrac{{2{\text{n}} - 1}}{{4{{\text{n}}^2}}}

Explanation

Solution

Here, we will use the concept of combination to solve the question. We have to divide the total number of boys into two equal groups. We will then subtract 1 boy from each group so that we separate the two tall boys and calculate the ways to divide remaining boys. Then we will use the formula of combination to find the answer.
Formula used: We will use the formula of combination, nCr=n!r!(n−r)!^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n}} - {\text{r}}} \right)!}}, where r{\text{r}} is the number of objects selected from n{\text{n}} number of set.

Complete step by step solution:
It is given that 2n boys are divided into two equal groups, that means there will be n boys in each group.
So that the total ways of dividing 2n boys will be 2nCn⋅nCn^{2{\text{n}}}{{\text{C}}_{\text{n}}}{ \cdot ^{\text{n}}}{{\text{C}}_{\text{n}}}.
As we have to find the probability of two tall boys, so we will leave two boys and find the number of ways to divide remaining boys.
After leaving the tall boys there will be 2n−22{\text{n}} - 2 boys which are left to be divided in the subgroups.
The number of ways in which 2n−22{\text{n}} - 2 boys can be divided from the 2 subgroups will be given by 2n - 2Cn⋅n - 1Cn^{2{\text{n - 2}}}{{\text{C}}_{\text{n}}}{ \cdot ^{{\text{n - 1}}}}{{\text{C}}_{\text{n}}}.
The two tall boys can be divided in two groups in 2 ways.
Now we will divide the two tall boys in different groups in 2n−2Cn−1⋅n−1Cn−1⋅22nCn⋅nCn\dfrac{{^{{\text{2n}} - {\text{2}}}{{\text{C}}_{{\text{n}} - {\text{1}}}}{ \cdot ^{{\text{n}} - {\text{1}}}}{{\text{C}}_{{\text{n}} - {\text{1}}}} \cdot 2}}{{^{{\text{2n}}}{{\text{C}}_{\text{n}}}{ \cdot ^{\text{n}}}{{\text{C}}_{\text{n}}}}} ways.
Now we will solve the above expression using the formula of combination, nCr=n!r!(n−r)!^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n}} - {\text{r}}} \right)!}}.

2n−2Cn−1⋅n−1Cn−1⋅22nCn⋅nCn=(2n−2)!(n−1)!(2n−2−(n−1))!⋅(n−1)!(n−1)!(n−1−(n−1))!⋅2(2n)!n!(2n−n)!⋅(n)!(n)!(n−n)! =(2n−2)!(n−1)!(n−1)!⋅10!⋅2(2n)!n!(n)!⋅10!\begin{array}{l}\dfrac{{^{{\text{2n}} - {\text{2}}}{{\text{C}}_{{\text{n}} - {\text{1}}}}{ \cdot ^{{\text{n}} - {\text{1}}}}{{\text{C}}_{{\text{n}} - {\text{1}}}} \cdot 2}}{{^{{\text{2n}}}{{\text{C}}_{\text{n}}}{ \cdot ^{\text{n}}}{{\text{C}}_{\text{n}}}}} = \dfrac{{\dfrac{{\left( {2{\text{n}} - 2} \right)!}}{{\left( {{\text{n}} - 1} \right)!\left( {2{\text{n}} - 2 - \left( {{\text{n}} - 1} \right)} \right)!}} \cdot \dfrac{{\left( {{\text{n}} - 1} \right)!}}{{\left( {{\text{n}} - 1} \right)!\left( {{\text{n}} - 1 - \left( {{\text{n}} - 1} \right)} \right)!}} \cdot 2}}{{\dfrac{{\left( {2{\text{n}}} \right)!}}{{{\text{n}}!\left( {2{\text{n}} - {\text{n}}} \right)!}} \cdot \dfrac{{\left( {\text{n}} \right)!}}{{\left( {\text{n}} \right)!\left( {{\text{n}} - {\text{n}}} \right)!}}}}\\\ = \dfrac{{\dfrac{{\left( {2{\text{n}} - 2} \right)!}}{{\left( {{\text{n}} - 1} \right)!\left( {{\text{n}} - 1} \right)!}} \cdot \dfrac{1}{{0!}} \cdot 2}}{{\dfrac{{\left( {2{\text{n}}} \right)!}}{{{\text{n}}!\left( {\text{n}} \right)!}} \cdot \dfrac{1}{{0!}}}}\end{array}
Simplifying the above equation, we get
2n−2Cn−1⋅n−1Cn−1⋅22nCn⋅nCn=(2n−2)!(n−1)!(n−1)!⋅1⋅2(2n)!n!(n)!⋅1 =(2n−2)!(n−1)!(n−1)!⋅1⋅2(2n)(2n−1)(2n−2)!n(n−1)!(n)(n−1)!⋅1\begin{array}{l}\dfrac{{^{{\text{2n}} - {\text{2}}}{{\text{C}}_{{\text{n}} - {\text{1}}}}{ \cdot ^{{\text{n}} - {\text{1}}}}{{\text{C}}_{{\text{n}} - {\text{1}}}} \cdot 2}}{{^{{\text{2n}}}{{\text{C}}_{\text{n}}}{ \cdot ^{\text{n}}}{{\text{C}}_{\text{n}}}}} = \dfrac{{\dfrac{{\left( {2{\text{n}} - 2} \right)!}}{{\left( {{\text{n}} - 1} \right)!\left( {{\text{n}} - 1} \right)!}} \cdot 1 \cdot 2}}{{\dfrac{{\left( {2{\text{n}}} \right)!}}{{{\text{n}}!\left( {\text{n}} \right)!}} \cdot 1}}\\\ = \dfrac{{\dfrac{{\left( {2{\text{n}} - 2} \right)!}}{{\left( {{\text{n}} - 1} \right)!\left( {{\text{n}} - 1} \right)!}} \cdot 1 \cdot 2}}{{\dfrac{{\left( {2{\text{n}}} \right)\left( {2{\text{n}} - 1} \right)\left( {2{\text{n}} - 2} \right)!}}{{{\text{n}}\left( {{\text{n}} - 1} \right)!\left( {\text{n}} \right)\left( {{\text{n}} - 1} \right)!}} \cdot 1}}\end{array}
Again simplifying the equation, we get
2n−2Cn−1⋅n−1Cn−1⋅22nCn⋅nCn=2⋅n⋅n(2n)(2n−1) =n(2n−1)\begin{array}{l}\dfrac{{^{{\text{2n}} - {\text{2}}}{{\text{C}}_{{\text{n}} - {\text{1}}}}{ \cdot ^{{\text{n}} - {\text{1}}}}{{\text{C}}_{{\text{n}} - {\text{1}}}} \cdot 2}}{{^{{\text{2n}}}{{\text{C}}_{\text{n}}}{ \cdot ^{\text{n}}}{{\text{C}}_{\text{n}}}}} = \dfrac{{2 \cdot {\text{n}} \cdot {\text{n}}}}{{\left( {2{\text{n}}} \right)\left( {2{\text{n}} - 1} \right)}}\\\ = \dfrac{{\text{n}}}{{\left( {2{\text{n}} - 1} \right)}}\end{array}

∴\therefore The correct answer is option A.

Note:
Here, we might make a mistake when calculating the factorial. It is important for us to understand that first we need to separate the tall boys to find the ways to divide the rest of the boys. If we find the number of ways to divide boys without separating tall boys, so there is a possibility that they might fall in the same group. So, we will not get the desired answer.