Question

A number of ways in which 2 Indian, 3 American, 3 Italian & 4 French men can be seated on a circle if the people of the same motional sit together is. -
A. $2!{(4!)^2}{\left( {3!} \right)^2}$
B.$2!{\text{ }}{\left( {3!} \right)^3}.4!$
C. $2!\left( {3!} \right){(4!)^3}$
D. none of these.

Answer 1

Hint: Here, at first find the man sitting together in a circle, then use this binomial formula on it. $\left( {n - 1} \right)!p!q!r!s!$ where \[\left( {n - 1} \right)\] is no. of arrangement of nations can be done, & $p!q!r!s!$ are for the no. of arrangements of countries can be done.

Complete step-by-step answer:
We know, no. of Indian(p) = 2
No. of American(q) = 3
No. of Italian (r)= 3
No. of French man(s) = 4
they can be seated in a circle if the people of the same motional sit together.
Here no. of arrangements of nations = \[\left( {n - 1} \right)\]= $4 - 1 = 3$ man to sit together in a circle.
therefore,
for Indian –${\text{2!}}$
for American – ${\text{3!}}$
for Italian – ${\text{3!}}$
for French man – ${\text{4!}}$
Now, we use the formula $\left( {n - 1} \right)!p!q!r!s!$
$ \Rightarrow 3!{\text{ 2! 3! 3! 4!}}$
$ \Rightarrow 2!{\text{ }}{\left( {3!} \right)^3}.4!$

So, the correct answer is “Option B”.

Additional information: the main concept one should have to solve this problem is of binomial formula , i.e $\left( {n - 1} \right)!p!q!r!s!$
Numerical related to sitting arrangements can be done using this formula -$\left( {n - 1} \right)!p!q!r!s!$

Note: We can use this formula $\left( {n - 1} \right)!p!q!r!s!$ for this type of problem, then find the answer. Do the calculations carefully so that there is no chance of minor error while putting the values of different terms.