7 boys and 5 girls are to be seated around a circular table such that no two girls sit together is?
7 boys and 5 girls are to be seated around a circular table such that no two girls sit together is?
Show Hint
- \(126(5!)^{2}\)
- \(720(5!)\)
- \(720(6!)\)
- 720
The Correct Option is A
Approach Solution - 1
The correct answer is (A) :
First, we need to find the total number of ways to seat all 12 people around the circular table, which is (12-1)! = 11! since we can fix one person's position as a reference.
Next, we need to subtract the number of ways that two or more girls sit together. We can approach this by treating the five girls as a block and permuting them first, which can be done in 5! ways.
Then we can insert this block of girls in the 8 spaces between the 7 boys or at the beginning or end of the line of boys, which gives us 9 positions to place the block of girls. Once the block of girls is placed, we can permute the 7 boys in 7! ways. Therefore, the total number of ways that two or more girls sit together is 5! × 9 × 7!
\(\therefore\) the number of ways that no two girls sit together is 11! - 5! × 9 × 7! = 126(5!)2.
Approach Solution -2
The correct answer is (A) : \(126(5!)^2\)
B1 , B2 , B3 , B4 , B5 , B6 , B7
Boys can be seated in (7 – 1)! ways = 6!
Now ways in which no two girls can be seated together is
\(6!\times^7C_5\times5!\)
\(6!\times \frac{7!}{5!2!}\times5!\)
\(=126(5!)^2\)
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