The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by a!× b! , then (a/b) is

Question

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by a!× b! , then (a/b) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

No. of ways in which 6 men can be arranged at a round table =(6-1)!=5! Now women can be arranged in 6P5 =6! ways Total Number of ways =6!× 5!= a! × b! ⇒ (a/b)=(6/5)=1.2

Q. The number of ways in which $6$ men and $5$ women can dine at a round table if no two women are to sit together is given by $a! \times b!$ , then $\frac{a}{b}$ is

No. of ways in which $6$ men can be arranged at a round table $= (6 - 1)! = 5!$
Now women can be arranged in $^{6} P_{5}$
$= 6!$ ways
Total Number of ways $= 6! \times 5! = a! \times b!$
$\Rightarrow \frac{a}{b} = \frac{6}{5} = 1.2$

Q. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by a!×b!, then ba​ is

No. of ways in which 6 men can be arranged at a round table =(6−1)!=5! Now women can be arranged in 6P5​ =6! ways Total Number of ways =6!×5!=a!×b! ⇒ba​=56​=1.2

Q. The number of ways in which $6$ men and $5$ women can dine at a round table if no two women are to sit together is given by $a! \times b!$ , then $\frac{a}{b}$ is

No. of ways in which $6$ men can be arranged at a round table $= (6 - 1)! = 5!$
Now women can be arranged in $^{6} P_{5}$
$= 6!$ ways
Total Number of ways $= 6! \times 5! = a! \times b!$
$\Rightarrow \frac{a}{b} = \frac{6}{5} = 1.2$