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Question: A number of ways in which 2 Indian, 3 American, 3 Italian & 4 French men can be seated on a circle i...

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(3!)22!{(4!)^2}{\left( {3!} \right)^2}
B.2! (3!)3.4!2!{\text{ }}{\left( {3!} \right)^3}.4!
C. 2!(3!)(4!)32!\left( {3!} \right){(4!)^3}
D. none of these.

Explanation

Solution

Here, at first find the man sitting together in a circle, then use this binomial formula on it. (n−1)!p!q!r!s!\left( {n - 1} \right)!p!q!r!s! where (n−1)\left( {n - 1} \right) is no. of arrangement of nations can be done, & p!q!r!s!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 = (n−1)\left( {n - 1} \right)= 4−1=34 - 1 = 3 man to sit together in a circle.
therefore,
for Indian –2!{\text{2!}}
for American – 3!{\text{3!}}
for Italian – 3!{\text{3!}}
for French man – 4!{\text{4!}}
Now, we use the formula (n−1)!p!q!r!s!\left( {n - 1} \right)!p!q!r!s!
⇒3! 2! 3! 3! 4!\Rightarrow 3!{\text{ 2! 3! 3! 4!}}
⇒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 (n−1)!p!q!r!s!\left( {n - 1} \right)!p!q!r!s!
Numerical related to sitting arrangements can be done using this formula -(n−1)!p!q!r!s!\left( {n - 1} \right)!p!q!r!s!

Note: We can use this formula (n−1)!p!q!r!s!\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.