Courses
Courses for Kids
Free study material
Free LIVE classes
More
Questions & Answers

# A tea party is arranged for 16 people along two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and two on the other side, in how many ways can they be seated?

Last updated date: 20th Mar 2023
Total views: 304.8k
Views today: 4.83k
Answer
Verified
304.8k+ views
Hint – In this question we have in total 16 seats with 8 on each side of the table. Firstly make these 4+2=6 people sit who have special demand in sitting arrangements. Now when these 6 people are taken care of we are left with 10 people and 4 seats on one side and 6 seats on the other side, use this concept to get the answer.

Complete step-by-step answer:
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats)
The number of ways of choosing 6 people out of 10 are ${}^{10}{C_6}$, now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are ${}^4{C_4} = 1$.
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is$\left( {8! \times 8!} \right)$.
So, a possible number of arrangements will be $\Rightarrow {}^{10}{C_6} \times {}^4{C_4} \times 8! \times 8!$
Now as we know${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
So, ${}^{10}{C_6} = \dfrac{{10!}}{{4! \times 6!}} = \dfrac{{10.9.8.7.6!}}{{4.3.2.1.6!}} = 210$
And ${}^4{C_4} = \dfrac{{4!}}{{4! \times 0!}} = 1$
So, total number of arrangements is
$= 210 \times \left( {8! \times 8!} \right)$.
So, this is the required answer.

Note – Whenever we face such types of problems the key point is to make special arrangements for the people who are in need of it, then arrange the remaining. Now combination comes with permutation as there are possibilities of these 8 people sitting on one side to rearrange. Thus this concept into consideration, to get through the answer.