# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. In how many ways can $15$ people be seated around two round tables with seating capacities of $7$ and $8$ people?
 ✖ A. $\dfrac{15!}{8!}$ ✖ B. $7!×8!$ ✔ C. ${^{15}C_8}×6!×7!$ ✖ D. $2×{^{15}C_7}×6!×7!$ ✖ E. ${^{15}C_8} × 8!$

Solution:
Option(C) is correct

Circular Permutation

$n$ objects can be arranged around a circle in $(n - 1)!$

If arranging these $n$ objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.

i.e., number of arrangements $=\dfrac{(n-1)!}{2}$

You can choose the $7$ people to sit in the first table in ${^{15}C_7}$ ways.

After selecting $7$ people for the table that can seat $7$ people, they can be seated in:

$(7-1)! = 6!$

The remaining $8$ people can be made to sit around the second circular table in:

$(8-1)! = 7!$ Ways.

Hence, total number of ways,

${^{15}C_8}× 6! × 7!$

## (5) Comment(s)

Ranger
()

Should we not select the table among the two tables first,which can be done in two ways ?

Daksh
()

yesss right answer should be option d

Anish
()

In the procedure it is written 7 people can be chosen in 15c7 ways. but in the answer choices the option is given as 15c8. the same thing has been done in the soln. why?

Abhishek Tripathi
()

because 15c7 and 15c8 is same nCr=nC(n-r)

Debasish Dey
()

Why n objects can be arranged around a circle in (n−1)! shouldn't it be n! ways?