(n-1)! ways
5 people, so 4! ways
I do not believe that answer is correct. Look at it this way:
Let the first person sit anywhere. Then the remaining 4 people can be seated in 4! (4 factorial = 4 * 3 * 2 * 1) Therefore 4! = 24 ways of seating 5 people around a circular table.
There are 27 choose 25 ways, which is equivalent to 27! / (25!(27-25)!) = 351 possible different groups of 25 people that can be formed from a total of 27 people.
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In how many ways can five children sit at round table?
yes
in how many ways can a party of 8 people arrange themselves round a table?
The number of circular permutations of n objects is (n-1)! so in this case the answer is 7! =5040 way to sit 8 people around a table.
7!(6*2*5*4*3*2*1)
6
7!
720
5! * 7!
When seating 4 Knights at 4 empty seats around a round table, we can fix one Knight in one seat to eliminate the rotations, effectively reducing the problem to arranging the remaining 3 Knights in the remaining 3 seats. The number of ways to arrange 3 Knights is given by 3! (3 factorial), which equals 6. Thus, there are 6 different ways for the 4 Knights to sit at the Round Table.
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To seat 5 men and 5 women alternately at a round table, we can fix one person to eliminate symmetrical arrangements. This leaves us with 4 men and 5 women to arrange. The men can be arranged in (4!) ways and the women in (5!) ways. Therefore, the total number of arrangements is (4! \times 5! = 24 \times 120 = 2880).