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Question Number 149955 Answers: 1 Comments: 0

If a group consist of 8 men and 6 women, in how many ways can a committee of 5 be selected if: i) the committee is to consist of 3 men and 3 women. ii) there are no restrictions on the number of men and women on the committee. iii) at least one man

$$\mathrm{If}\:\mathrm{a}\:\mathrm{group}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{8}\:\mathrm{men}\:\mathrm{and}\:\mathrm{6}\:\mathrm{women}, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{committee}\:\mathrm{of} \\ $$$$\mathrm{5}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{if}: \\ $$$$\left.\:\:\:\:\:\:\mathrm{i}\right)\:\mathrm{the}\:\mathrm{committee}\:\mathrm{is}\:\mathrm{to}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{3}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{3}\:\mathrm{women}. \\ $$$$\left.\:\:\:\:\:\:\mathrm{ii}\right)\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{restrictions}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{and}\:\mathrm{women}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{committee}. \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{iii}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{man} \\ $$

Question Number 149817 Answers: 0 Comments: 0

x^3 −x=c let x=t+h t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c=0 let t(t^2 +3h^2 −1)=p 3ht^2 +h^3 −h−c=q ⇒ t^2 =((q+c+h−h^3 )/(3h)) p+q=0 (((q+c+h−h^3 )/(3h)))(((q+c+h−h^3 )/(3h))+3h^2 −1)^2 =q^2 ⇒ (q+c+h−h^3 )(q+c−2h+8h^3 )^2 = 27h^3 q^2 let q+c−2h=0 ⇒ 64h^4 (3−h^2 )=27(2h−c)^2 ⇒ 8h^2 (√(3−h^2 ))=3(√3)(2h−c) let h=(√3)sin θ ⇒ 8sin^2 θcos θ=2(√3)sin θ−c 4sin θsin 2θ=2(√3)sin θ−c 2sin θ((√3)−2sin 2θ)=c ......

$$\:\:\:{x}^{\mathrm{3}} −{x}={c} \\ $$$$\:\:{let}\:\:{x}={t}+{h} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{t}\left({t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={p} \\ $$$$\mathrm{3}{ht}^{\mathrm{2}} +{h}^{\mathrm{3}} −{h}−{c}={q} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} =\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}} \\ $$$${p}+{q}=\mathrm{0} \\ $$$$\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}\right)\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:={q}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({q}+{c}+{h}−{h}^{\mathrm{3}} \right)\left({q}+{c}−\mathrm{2}{h}+\mathrm{8}{h}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{27}{h}^{\mathrm{3}} {q}^{\mathrm{2}} \\ $$$${let}\:\:{q}+{c}−\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{64}{h}^{\mathrm{4}} \left(\mathrm{3}−{h}^{\mathrm{2}} \right)=\mathrm{27}\left(\mathrm{2}{h}−{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{2}} \sqrt{\mathrm{3}−{h}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{3}}\left(\mathrm{2}{h}−{c}\right) \\ $$$${let}\:\:{h}=\sqrt{\mathrm{3}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{8sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{4sin}\:\theta\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{2sin}\:\theta\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\mathrm{2}\theta\right)={c} \\ $$$$...... \\ $$

Question Number 149813 Answers: 1 Comments: 1

∫(−1)^([x]) dx = ?

$$\int\left(−\mathrm{1}\right)^{\left[\boldsymbol{\mathrm{x}}\right]} \:\mathrm{dx}\:=\:? \\ $$

Question Number 149809 Answers: 2 Comments: 0