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Home  >>  EAMCET  >>  Mathematics
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10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is :

\[\begin {array} {1 1} (a)\;11!\;10! & \quad (b)\;\frac{11 !}{6!\;5!} \\ (c)\;\frac{10 !\; 9!}{5!} & \quad (d)\;\frac{11 !\; 10 !}{5 !} \end {array}\]

Can you answer this question?
 
 

1 Answer

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$(d)\;\large\frac{11 !\; 10 !}{5 !}$
answered Nov 7, 2013 by pady_1
 

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