There are 10 person among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers , is equal to:
Answers
Answer:
The total number of ways is 7! x 2!.
Step-by-step explanation:
Since there are 10 person among whom two are brother, we want to know the total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers.
In all cases we want on person to stay between the brothers, so we can count that as 1 person, then we want the two brothers to stay one seat away, and they can permute between themselves, then we count 2 x 1, i.e, 2!.
Hence, by now we have 1 x 2! = 2!.
Then the remaining 7 people can permute with no condition, then we get 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7!
Thus, the total number of ways is 7! x 2!.
Answer:
8! * 2!.
Step-by-step explanation:
Total number of persons = 10.
Total number of brothers without 2 brothers = 10 – 2 = 8
Fix the 2 brothers position as B1 _ B2.
Both brothers can be fixed in 2! Ways.
The person in between can be any of 8 remaining persons. (8 ways).
Consider 2 brothers and in between person as one block.
This block probability = 8 * 2!
Now consider remaining 7 persons and this block (Total 8 persons or entities).
n people can sit around a table in (n – 1)! Combinations.
Remaining 8 entities can be arranged in 7! Ways.
So total combinations are 2! * 8 * 7! = 8! * 2!.