7 boys and 5 girls are to be seated around a circular table such that no two girls sit together is?
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.
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
\(6!\times^7C_5\times5!\)
\(6!\times \frac{7!}{5!2!}\times5!\)
\(=126(5!)^2\)
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:
Permutation is the method or the act of arranging members of a set into an order or a sequence.
Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.