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Question Number 135490 by benjo_mathlover last updated on 13/Mar/21

Combination

$${Combination} \\ $$ A committee of 8 people is to be formed from 7 women and 5 men. In how many ways can the members be chosen so as to include at least 3 men?\\n

Answered by EDWIN88 last updated on 13/Mar/21

= ((5),(3) ) ((7),(5) ) + ((5),(4) ) ((7),(4) ) + ((5),(3) ) ((7),(3) ) = 420

$$=\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{5}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{4}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{4}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{3}}\end{pmatrix}\:=\:\mathrm{420} \\ $$

Answered by mr W last updated on 13/Mar/21

the coef. of term x^8 in (C_3 ^5 x^3 +C_4 ^5 x^4 +C_5 ^5 x^5 )(1+C_1 ^7 x+C_2 ^7 x^2 +...+C_7 ^7 x^7 ) is 420. that means there are 420 ways.

$${the}\:{coef}.\:{of}\:{term}\:{x}^{\mathrm{8}} \:{in} \\ $$ $$\left({C}_{\mathrm{3}} ^{\mathrm{5}} {x}^{\mathrm{3}} +{C}_{\mathrm{4}} ^{\mathrm{5}} {x}^{\mathrm{4}} +{C}_{\mathrm{5}} ^{\mathrm{5}} {x}^{\mathrm{5}} \right)\left(\mathrm{1}+{C}_{\mathrm{1}} ^{\mathrm{7}} {x}+{C}_{\mathrm{2}} ^{\mathrm{7}} {x}^{\mathrm{2}} +...+{C}_{\mathrm{7}} ^{\mathrm{7}} {x}^{\mathrm{7}} \right) \\ $$ $${is}\:\mathrm{420}.\:{that}\:{means}\:{there}\:{are}\:\mathrm{420}\:{ways}. \\ $$

Commented bymr W last updated on 13/Mar/21

Answered by physicstutes last updated on 13/Mar/21

i like looking at it from this pespective. see the table below which shows the number of men and women selected based on the above conditions determinant (((number of men),(number of women)),(3,5),(4,4),(5,3)) therefore the number of ways = ^5 C_3 ×^8 C_5 + ^5 C_4 ×^8 C_4 + ^5 C_5 ×^5 C_3 = 420 ways

$$\mathrm{i}\:\mathrm{like}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{it}\:\mathrm{from}\:\mathrm{this}\:\mathrm{pespective}.\:\mathrm{see}\:\mathrm{the}\:\mathrm{table}\:\mathrm{below} \\ $$ $$\mathrm{which}\:\mathrm{shows}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{and}\:\mathrm{women}\:\mathrm{selected}\:\mathrm{based}\:\mathrm{on} \\ $$ $$\mathrm{the}\:\mathrm{above}\:\mathrm{conditions} \\ $$ $$\:\begin{array}{|c|c|c|c|}{\mathrm{number}\:\mathrm{of}\:\mathrm{men}}&\hline{\mathrm{number}\:\mathrm{of}\:\mathrm{women}}\\{\mathrm{3}}&\hline{\mathrm{5}}\\{\mathrm{4}}&\hline{\mathrm{4}}\\{\mathrm{5}}&\hline{\mathrm{3}}\\\hline\end{array} \\ $$ $$\:\mathrm{therefore}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:=\:^{\mathrm{5}} {C}_{\mathrm{3}} ×^{\mathrm{8}} {C}_{\mathrm{5}} \:+\:^{\mathrm{5}} {C}_{\mathrm{4}} ×^{\mathrm{8}} {C}_{\mathrm{4}} \:+\:^{\mathrm{5}} {C}_{\mathrm{5}} ×^{\mathrm{5}} {C}_{\mathrm{3}} \:=\:\mathrm{420}\:\mathrm{ways} \\ $$

Commented bymr W last updated on 13/Mar/21

in this case you can use this method. but for complex cases for example there are 10 men, 12 women, 8 boys, 7 girls. a committee with 10 persons should be formed, we need other methods.

$${in}\:{this}\:{case}\:{you}\:{can}\:{use}\:{this}\:{method}. \\ $$ $${but}\:{for}\:{complex}\:{cases}\:{for}\:{example} \\ $$ $${there}\:{are}\:\mathrm{10}\:{men},\:\mathrm{12}\:{women},\:\mathrm{8}\:{boys}, \\ $$ $$\mathrm{7}\:{girls}.\:{a}\:{committee}\:{with}\:\mathrm{10}\:{persons} \\ $$ $${should}\:{be}\:{formed},\:{we}\:{need}\:{other} \\ $$ $${methods}. \\ $$

Commented byphysicstutes last updated on 13/Mar/21

sure sir i totally know that.

$$\mathrm{sure}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{totally}\:\mathrm{know}\:\mathrm{that}. \\ $$

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