There are 7 persons which include a group of 3 friends F1, F2, F3. Number of ways they can be seated on a round table if atleast two out of F 1, F2, F3 are seated next to each other is

Question

There are 7 persons which include a group of 3 friends F1, F2, F3. Number of ways they can be seated on a round table if atleast two out of F 1, F2, F3 are seated next to each other is (A) 144 (B) 576 (C) 360 (D) 432

Tardigrade Admin · Accepted Answer

3 ! × 4 C3 × 3 !=6 × 4 × 6 = 36 × 4 = 14

∴ Required ways = Total - 144 = 720 - 144 = 576

Q. There are $7$ persons which include a group of $3$ friends $F_{1}, F_{2}, F_{3}$ . Number of ways they can be seated on a round table if atleast two out of $F_{1}$ , $F_{2}, F_{3}$ are seated next to each other is

144

576

360

432

$3! \times^{4} C_{3} \times 3! = 6 \times 4 \times 6$
$= 36 \times 4 = 14$

$∴$ Required ways = Total $- 144$
$= 720 - 144 = 576$

Q. There are 7 persons which include a group of 3 friends F1​,F2​,F3​. Number of ways they can be seated on a round table if atleast two out of F1​, F2​,F3​ are seated next to each other is

144

576

360

432

3!×4C3​×3!=6×4×6 =36×4=14 ∴ Required ways = Total −144 =720−144=576

Q. There are $7$ persons which include a group of $3$ friends $F_{1}, F_{2}, F_{3}$ . Number of ways they can be seated on a round table if atleast two out of $F_{1}$ , $F_{2}, F_{3}$ are seated next to each other is

$3! \times^{4} C_{3} \times 3! = 6 \times 4 \times 6$
$= 36 \times 4 = 14$

$∴$ Required ways = Total $- 144$
$= 720 - 144 = 576$