
In how many distinct ways can \[3\] ladies and \[3\] gentlemen be seated at a round table, so that exactly any two ladies sit together?
Answer
465.3k+ views
Hint: Here, we have to find the ways in which \[3\] ladies and \[3\] gentlemen are seated at a round table, so that exactly any two ladies sit together. In order to solve this question we will use the concept of permutation and combination. Permutation and combination can be defined as the methods of counting which help us to determine the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them.
Complete step by step answer:
According to the given question there are \[3\] ladies and \[3\] gentlemen and we have to find ways that they can be seated at a round table in such a way that exactly any two ladies sit together. So, we will use the concept of permutation and combination. Permutation is defined as the arrangement of objects in a definite order whereas combination is defined as the selection of some or all objects from a given set of different objects when order of selection is not considered.
Firstly, we will find the number of ways in which any two ladies can be sit together so we will use the formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. So,
${ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{2!(3 - 2)!}}$
Simplifying the denominator we get,
${ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{2!\,\, \times \,\,1!}}$
${ \Rightarrow ^3}{C_2} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}}$
On dividing the numbers we get,
${ \Rightarrow ^3}{C_2} = 3$
Now there are $4$ members remaining so they can be arranged in $4$ different ways i.e., in $4!$ ways. Therefore the total numbers of ways in which \[3\] ladies and \[3\] gentleman can be seated at a round table, so that exactly any two ladies sit together can be calculated as,
$ \Rightarrow 4!\, \times {\,^3}{C_2}$
Putting the value of $^3{C_2} = 3$ we get,
$ \Rightarrow 4!\,\, \times 3$
$ \Rightarrow 4\, \times 3 \times 2 \times 1 \times 3\,$
On multiplying the numbers we get,
$ \therefore 24 \times 2 = 72$
Hence, in $72$ the number of ways \[3\] ladies and \[3\] gentleman can be seated at a round table, so that exactly any two ladies sit together.
Note: There is a difference between permutation and combination. We can define permutation as the number of ways to arrange objects and combination is defined as the number of ways to select objects. In permutation order is to be considered and in combination we do not consider order.
Complete step by step answer:
According to the given question there are \[3\] ladies and \[3\] gentlemen and we have to find ways that they can be seated at a round table in such a way that exactly any two ladies sit together. So, we will use the concept of permutation and combination. Permutation is defined as the arrangement of objects in a definite order whereas combination is defined as the selection of some or all objects from a given set of different objects when order of selection is not considered.
Firstly, we will find the number of ways in which any two ladies can be sit together so we will use the formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. So,
${ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{2!(3 - 2)!}}$
Simplifying the denominator we get,
${ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{2!\,\, \times \,\,1!}}$
${ \Rightarrow ^3}{C_2} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}}$
On dividing the numbers we get,
${ \Rightarrow ^3}{C_2} = 3$
Now there are $4$ members remaining so they can be arranged in $4$ different ways i.e., in $4!$ ways. Therefore the total numbers of ways in which \[3\] ladies and \[3\] gentleman can be seated at a round table, so that exactly any two ladies sit together can be calculated as,
$ \Rightarrow 4!\, \times {\,^3}{C_2}$
Putting the value of $^3{C_2} = 3$ we get,
$ \Rightarrow 4!\,\, \times 3$
$ \Rightarrow 4\, \times 3 \times 2 \times 1 \times 3\,$
On multiplying the numbers we get,
$ \therefore 24 \times 2 = 72$
Hence, in $72$ the number of ways \[3\] ladies and \[3\] gentleman can be seated at a round table, so that exactly any two ladies sit together.
Note: There is a difference between permutation and combination. We can define permutation as the number of ways to arrange objects and combination is defined as the number of ways to select objects. In permutation order is to be considered and in combination we do not consider order.
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