Question

Eight men and six women sit around a table. How many of sitting arrangements will have no
two women together?

Answer 1

Given Question -

• Eight men and six women sit around a table. How many of sitting arrangements will have no two women together?

ANSWER

CONCEPT USED -

CONCEPT USED -CIRCULAR PERMUTATION

• The number of ways to arrange 'n' distinct objects along a fixed circle is (n - 1)!.

PERMUTATION

If there are 'n' distinct objects, out of which 'r' objects have to be arranged, the number of such arrangement is given by

$\bigstar \: \: \boxed{\tt \:^{n}P_r=\dfrac{n!}{(n-r)!}}$

$\large\underline\purple{\bold{Solution :- }}$

• The number of ways in which 8 men can be arranged themselves around a circular table is given by (8 - 1)! = 7!.

Now,

• We have to arrange 6 women in between these mens so that no two women sit together.

• Now, there are 8 gaps between these 8 men around a round table.

Now,

• Number of ways in which 6 women can arrange themselves in 8 seats is given by

$\tt \: number \: of \: ways \: = \:^{8}P_6$

$\tt \: number \: of \: ways \: =\dfrac{8!}{(8 - 6)!}$

$\tt \: number \: of \: ways \: =\dfrac{8!}{2!}$

$\tt \: number \: of \: ways \: =\dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}$

$\tt \: number \: of \: ways \: = \: 20160$

Hence,

• Number of sitting arrangements in which 8 men and 6 women can be arranged around a round table so that no two women sit together is

$\rm :\implies\:\tt \: number \: of \: ways \: =20160 \times 7!$

OR

$\rm :\implies\:\tt \: number \: of \: ways \: = \: \:^{8}P_6 \: \times 7!$