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Question Number 123771 by benjo_mathlover last updated on 28/Nov/20

Commented by benjo_mathlover last updated on 28/Nov/20

let ⧫ = man , ♥=woman ⧫♥⧫♥⧫♥⧫♥⧫♥⧫♥⧫= 3! ×C_3 ^6 ×5! ×42 p(A)= ((3!×C_3 ^6 ×7!)/(10!)) = ((6×20×7!)/(10×9×8×7!))=(1/6)

$$\:{let}\:\blacklozenge\:=\:{man}\:,\:\heartsuit={woman} \\ $$$$\:\blacklozenge\heartsuit\blacklozenge\heartsuit\blacklozenge\heartsuit\blacklozenge\heartsuit\blacklozenge\heartsuit\blacklozenge\heartsuit\blacklozenge= \\ $$$$\:\mathrm{3}!\:×{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{5}!\:×\mathrm{42} \\ $$$${p}\left({A}\right)=\:\frac{\mathrm{3}!×{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{7}!}{\mathrm{10}!}\:=\:\frac{\mathrm{6}×\mathrm{20}×\mathrm{7}!}{\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}!}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\: \\ $$

Commented by benjo_mathlover last updated on 28/Nov/20

it is correct ?

$${it}\:{is}\:{correct}\:? \\ $$

Commented by mr W last updated on 28/Nov/20

correct! but what did you think by writing 5!×42? to place three women into the six places between the 7 men there are C_3 ^6 ways, to arrange the 3 women and 7 men there are 3!×7! ways, therefore the answer is p(A)=((C_3 ^6 ×3!×7!)/(10!))

$${correct}!\:{but}\:{what}\:{did}\:{you}\:{think}\:{by} \\ $$$${writing}\:\mathrm{5}!×\mathrm{42}? \\ $$$${to}\:{place}\:{three}\:{women}\:{into}\:{the}\:{six}\: \\ $$$${places}\:{between}\:{the}\:\mathrm{7}\:{men}\:{there}\:{are} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{6}} \:{ways},\:{to}\:{arrange}\:{the}\:\mathrm{3}\:{women}\:{and} \\ $$$$\mathrm{7}\:{men}\:{there}\:{are}\:\mathrm{3}!×\mathrm{7}!\:{ways},\: \\ $$$${therefore}\:{the}\:{answer}\:{is} \\ $$$${p}\left({A}\right)=\frac{{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!×\mathrm{7}!}{\mathrm{10}!} \\ $$

Commented by benjo_mathlover last updated on 28/Nov/20

because at both ends have to be occupied by men then many ways 6x7 = 42 sir

Commented by mr W last updated on 28/Nov/20

but we don′t need to arrange the two men at ends at first and then the other five men, we can arrange all 7 men all at once. certainly the result is the same: 7×6×5!=7!

$${but}\:{we}\:{don}'{t}\:{need}\:{to}\:{arrange}\:{the}\:{two} \\ $$$${men}\:{at}\:{ends}\:{at}\:{first}\:{and}\:{then}\:{the} \\ $$$${other}\:{five}\:{men},\:{we}\:{can}\:{arrange}\:{all} \\ $$$$\mathrm{7}\:{men}\:{all}\:{at}\:{once}.\:{certainly}\:{the} \\ $$$${result}\:{is}\:{the}\:{same}: \\ $$$$\mathrm{7}×\mathrm{6}×\mathrm{5}!=\mathrm{7}! \\ $$

Commented by benjo_mathlover last updated on 28/Nov/20

oo thank you sir

$${oo}\:{thank}\:{you}\:{sir} \\ $$

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