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Question Number 120028 by Don08q last updated on 28/Oct/20

Answered by bramlexs22 last updated on 29/Oct/20

(a) say three particular women is { (w_1 ),(w_2 ),(w_3 ) :} ■=woman particular □= the other person □■□■□■□■□⇒C _3^5 ×C _3^4 ×3!×5! = 10×3×6×120 =180×120=21,600

$$\left({a}\right)\:{say}\:{three}\:{particular}\:{women}\:{is}\:\begin{cases}{{w}_{\mathrm{1}} }\\{{w}_{\mathrm{2}} }\\{{w}_{\mathrm{3}} }\end{cases} \\ $$$$\blacksquare={woman}\:{particular} \\ $$$$\Box=\:{the}\:{other}\:{person} \\ $$$$\Box\blacksquare\Box\blacksquare\Box\blacksquare\Box\blacksquare\Box\Rightarrow{C}\:_{\mathrm{3}} ^{\mathrm{5}} ×{C}\:_{\mathrm{3}} ^{\mathrm{4}} ×\mathrm{3}!×\mathrm{5}!\: \\ $$$$=\:\mathrm{10}×\mathrm{3}×\mathrm{6}×\mathrm{120} \\ $$$$=\mathrm{180}×\mathrm{120}=\mathrm{21},\mathrm{600} \\ $$$$ \\ $$

Answered by mr W last updated on 29/Oct/20

(a) □M□M□M□W□W□ □=place for three particular wemen C_3 ^6 ×3!×5!=14 400 (b) □M■M■M□ (1+x+x^2 +...)^2 (x+x+x^3 +...)^2 =(x^2 /((1−x)^4 ))=x^2 Σ_(k=0) ^∞ C_3 ^(k+3) x^k k=3, C_3 ^6 ⇒C_3 ^6 ×3!×5!=14 400 (c) (((8−1)!)/2)=((7!)/2)=2520

$$\left({a}\right) \\ $$$$\Box{M}\Box{M}\Box{M}\Box{W}\Box{W}\Box \\ $$$$\Box={place}\:{for}\:{three}\:{particular}\:{wemen} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!×\mathrm{5}!=\mathrm{14}\:\mathrm{400} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\Box{M}\blacksquare{M}\blacksquare{M}\Box \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right)^{\mathrm{2}} \left({x}+{x}+{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }={x}^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${k}=\mathrm{3},\:{C}_{\mathrm{3}} ^{\mathrm{6}} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!×\mathrm{5}!=\mathrm{14}\:\mathrm{400} \\ $$$$ \\ $$$$\left({c}\right) \\ $$$$\frac{\left(\mathrm{8}−\mathrm{1}\right)!}{\mathrm{2}}=\frac{\mathrm{7}!}{\mathrm{2}}=\mathrm{2520} \\ $$

Commented by bramlexs22 last updated on 29/Oct/20

(a) how do you choice 3 woman particular from 5 woman ?

$$\left({a}\right)\:{how}\:{do}\:{you}\:{choice}\:\mathrm{3}\:{woman}\:{particular} \\ $$$${from}\:\mathrm{5}\:{woman}\:? \\ $$

Commented by mr W last updated on 29/Oct/20

they are three particular wowen. we should not choose them.

$${they}\:{are}\:{three}\:{particular}\:{wowen}. \\ $$$${we}\:{should}\:{not}\:{choose}\:{them}. \\ $$

Commented by bramlexs22 last updated on 29/Oct/20

oo it does meant the three woman particular fix ?

$${oo}\:{it}\:{does}\:{meant}\:{the}\:{three}\:{woman} \\ $$$${particular}\:{fix}\:? \\ $$

Commented by mr W last updated on 29/Oct/20

yes. let′s say the wemen are A,B,C,D,E. the three particular wemen are A,B,C. they are fixed.

$${yes}.\:{let}'{s}\:{say}\:{the}\:{wemen}\:{are}\:{A},{B},{C},{D},{E}. \\ $$$${the}\:{three}\:{particular}\:{wemen}\:{are} \\ $$$${A},{B},{C}.\:{they}\:{are}\:{fixed}. \\ $$

Commented by Don08q last updated on 29/Oct/20

Thank you Sir, but why ^6 C_3 and not ^6 P_3

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir},\:{but}\:{why}\:\:^{\mathrm{6}} \mathrm{C}_{\mathrm{3}} \:{and}\:{not}\:\:^{\mathrm{6}} \mathrm{P}_{\mathrm{3}} \: \\ $$

Commented by mr W last updated on 29/Oct/20

C_3 ^6 ×3!=P_3 ^6

$$\mathrm{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!=\mathrm{P}_{\mathrm{3}} ^{\mathrm{6}} \: \\ $$

Commented by Don08q last updated on 29/Oct/20

OK. Thank you

$$\:\mathrm{OK}.\:\mathrm{Thank}\:\mathrm{you} \\ $$

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