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Question Number 870 by 112358 last updated on 05/Apr/15
$${In}\:{how}\:{many}\:{ways}\:{may}\:{a}\:{line}\: \\ $$$${consisting}\:{of}\:\mathrm{4}\:{men}\:{and}\:\mathrm{4}\:{women} \\ $$$${be}\:{acquired}\:{if}\:{no}\:{two}\:{women}\:{are} \\ $$$${to}\:{be}\:{adjacent}\:{to}\:{each}\:{other}? \\ $$
Answered by prakash jain last updated on 05/Apr/15
$$\mathrm{A}.\:\mathrm{Let}\:\mathrm{women}\:\mathrm{take}\:\mathrm{position}\:\mathrm{1},\:\mathrm{3},\mathrm{5},\mathrm{7} \\ $$$$\mathrm{and}\:\mathrm{men}\:\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}. \\ $$$$\mathrm{Seating}\:\mathrm{arrangment}\:\mathrm{for}\:\mathrm{men}=\mathrm{4}! \\ $$$$\mathrm{Seating}\:\mathrm{arrangment}\:\mathrm{for}\:\mathrm{women}=\mathrm{4}! \\ $$$$\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{seating}\:\mathrm{arrangment}\:\mathrm{A}=\mathrm{4}!×\mathrm{4}! \\ $$$$\mathrm{B}.\:\mathrm{Let}\:\:\mathrm{men}\:\mathrm{take}\:\mathrm{position}\:\mathrm{1},\:\mathrm{3},\mathrm{5},\mathrm{7} \\ $$$$\mathrm{and}\:\mathrm{women}\:\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}. \\ $$$$\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{seating}\:\mathrm{arrangment}\:\mathrm{B}=\mathrm{4}!×\mathrm{4}! \\ $$$$\mathrm{Total}\:=\mathrm{A}+\mathrm{B}=\mathrm{2}×\mathrm{4}!×\mathrm{4}! \\ $$