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Question Number 97045 by john santu last updated on 06/Jun/20
a committee consisting of  5 men & 4 women is to be   chosen from 8 men & 4 women.  If one man & one woman are husband   & wife , how many ways can   the committee be chosen if  only one of the husband & wife  must be chosen ?
$$\mathrm{a}\:\mathrm{committee}\:\mathrm{consisting}\:\mathrm{of} \\ $$$$\mathrm{5}\:\mathrm{men}\:\&\:\mathrm{4}\:\mathrm{women}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{8}\:\mathrm{men}\:\&\:\mathrm{4}\:\mathrm{women}. \\ $$$$\mathrm{If}\:\mathrm{one}\:\mathrm{man}\:\&\:\mathrm{one}\:\mathrm{woman}\:\mathrm{are}\:\mathrm{husband}\: \\ $$$$\&\:\mathrm{wife}\:,\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\: \\ $$$$\mathrm{the}\:\mathrm{committee}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{if} \\ $$$$\mathrm{only}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{husband}\:\&\:\mathrm{wife} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{chosen}\:? \\ $$
Commented by bemath last updated on 06/Jun/20
⇔ C(7,5) = ((7.6.)/(2.1)) = 21
$$\Leftrightarrow\:\mathrm{C}\left(\mathrm{7},\mathrm{5}\right)\:=\:\frac{\mathrm{7}.\mathrm{6}.}{\mathrm{2}.\mathrm{1}}\:=\:\mathrm{21} \\ $$
Commented by mr W last updated on 06/Jun/20
since there are only 4 women and  4 women must be chosen, so all  women must be chosen and the wife  is always chosen and the husband  can never be chosen. so the question  is in how many ways can 5 men be  chosen from the other 7 men, the  answer is C_5 ^7 =21.
$${since}\:{there}\:{are}\:{only}\:\mathrm{4}\:{women}\:{and} \\ $$$$\mathrm{4}\:{women}\:{must}\:{be}\:{chosen},\:{so}\:{all} \\ $$$${women}\:{must}\:{be}\:{chosen}\:{and}\:{the}\:{wife} \\ $$$${is}\:{always}\:{chosen}\:{and}\:{the}\:{husband} \\ $$$${can}\:{never}\:{be}\:{chosen}.\:{so}\:{the}\:{question} \\ $$$${is}\:{in}\:{how}\:{many}\:{ways}\:{can}\:\mathrm{5}\:{men}\:{be} \\ $$$${chosen}\:{from}\:{the}\:{other}\:\mathrm{7}\:{men},\:{the} \\ $$$${answer}\:{is}\:{C}_{\mathrm{5}} ^{\mathrm{7}} =\mathrm{21}. \\ $$
Commented by bemath last updated on 06/Jun/20
oo sorry. i misread the question
$$\mathrm{oo}\:\mathrm{sorry}.\:\mathrm{i}\:\mathrm{misread}\:\mathrm{the}\:\mathrm{question}\: \\ $$
Commented by john santu last updated on 06/Jun/20
yes. right. thank you
$$\mathrm{yes}.\:\mathrm{right}.\:\mathrm{thank}\:\mathrm{you} \\ $$

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