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Question Number 135218 by benjo_mathlover last updated on 11/Mar/21

Permutation

$$\mathrm{Permutation} \\ $$ How many ways can 10 men and 7 women sit at a round table so that no 2 women are next to each other? ��������\\n

Answered by EDWIN88 last updated on 11/Mar/21

⇔   : place of woman, C_7 ^( 10)  = 120  the number of ways arrangement   is = 7!×9!×120

$$\Leftrightarrow\: \::\:\mathrm{place}\:\mathrm{of}\:\mathrm{woman},\:\mathrm{C}_{\mathrm{7}} ^{\:\mathrm{10}} \:=\:\mathrm{120} \\ $$ $$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{arrangement}\: \\ $$ $$\mathrm{is}\:=\:\mathrm{7}!×\mathrm{9}!×\mathrm{120}\: \\ $$

Commented byEDWIN88 last updated on 11/Mar/21

Commented byEDWIN88 last updated on 11/Mar/21

typo

$$\mathrm{typo} \\ $$

Answered by mr W last updated on 11/Mar/21

we place at first the 10 men. there  are 9! ways. then we place the 7  women in the 10 positions among  the men, there are C_7 ^(10) ×7! ways.  totally 9!×7!×C_7 ^(10)

$${we}\:{place}\:{at}\:{first}\:{the}\:\mathrm{10}\:{men}.\:{there} \\ $$ $${are}\:\mathrm{9}!\:{ways}.\:{then}\:{we}\:{place}\:{the}\:\mathrm{7} \\ $$ $${women}\:{in}\:{the}\:\mathrm{10}\:{positions}\:{among} \\ $$ $${the}\:{men},\:{there}\:{are}\:{C}_{\mathrm{7}} ^{\mathrm{10}} ×\mathrm{7}!\:{ways}. \\ $$ $${totally}\:\mathrm{\color{mathred}{9}}\color{mathred}{!}\color{mathred}{×}\mathrm{\color{mathred}{7}}\color{mathred}{!}\color{mathred}{×}{\color{mathred}{C}}_{\mathrm{\color{mathred}{7}}} ^{\mathrm{\color{mathred}{1}\color{mathred}{0}}} \\ $$

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