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Question-224623

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Question Number 224623 by gregori last updated on 22/Sep/25
$$\:\:\: \\ $$
Commented by gregori last updated on 23/Sep/25
no sir. i don′t get answer
$${no}\:{sir}.\:{i}\:{don}'{t}\:{get}\:{answer}\: \\ $$
Commented by mr W last updated on 22/Sep/25
8 467 200 arrangements?
$$\mathrm{8}\:\mathrm{467}\:\mathrm{200}\:{arrangements}? \\ $$
Answered by mehdee7396 last updated on 22/Sep/25
6!× ((7),(4) )×4!=((6!7!)/(3!))
$$\mathrm{6}!×\begin{pmatrix}{\mathrm{7}}\\{\mathrm{4}}\end{pmatrix}×\mathrm{4}!=\frac{\mathrm{6}!\mathrm{7}!}{\mathrm{3}!} \\ $$
Answered by mr W last updated on 24/Sep/25
6 men and 1 empty chair =7 men 4 women and 7 men sit in 11 chairs AWBWBWBWA A=zero or more men=1+x+x^2 +... B=one or more men=x+x^2 +x^3 +... (1+x+x^2 +x^3 +...)^2 (x+x^2 +x^3 +...)^3 =x^3 (1+x+x^2 +x^3 +...)^5 =(x^3 /((1−x)^5 ))=x^3 Σ_(k=0) ^∞ C_4 ^(k+4) x^k coef. of x^7 is C_4 ^8 ⇒answer is C_4 ^8 ×4!×7!=8 467 200 ✓
$$\mathrm{6}\:{men}\:{and}\:\mathrm{1}\:{empty}\:{chair}\:=\mathrm{7}\:{men} \\ $$$$\mathrm{4}\:{women}\:{and}\:\mathrm{7}\:{men}\:{sit}\:{in}\:\mathrm{11}\:{chairs} \\ $$$${AWBWBWBWA} \\ $$$${A}={zero}\:{or}\:{more}\:{men}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +… \\ $$$${B}={one}\:{or}\:{more}\:{men}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +… \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{2}} \left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{5}} \\ $$$$=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }={x}^{\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{7}} \:{is}\:{C}_{\mathrm{4}} ^{\mathrm{8}} \\ $$$$\Rightarrow{answer}\:{is}\:{C}_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{4}!×\mathrm{7}!=\mathrm{8}\:\mathrm{467}\:\mathrm{200}\:\checkmark \\ $$
Commented by mr W last updated on 24/Sep/25
Alternative method: arrange the 6 men and the empty chair (= 7 men): _M_M_M_M_M_M_M_ there are 7! ways to do this. now arrange the 4 women. each of them can take one of the 8 places “_”. there are C_4 ^8 ×4!=P_4 ^8 ways to do this. so the answer is 7!×P_4 ^8 =8 467 200.
$${Alternative}\:{method}: \\ $$$${arrange}\:{the}\:\mathrm{6}\:{men}\:{and}\:{the}\:{empty} \\ $$$${chair}\:\left(=\:\mathrm{7}\:{men}\right): \\ $$$$\_{M\_M\_M\_M\_M\_M\_M\_} \\ $$$${there}\:{are}\:\mathrm{7}!\:{ways}\:{to}\:{do}\:{this}. \\ $$$${now}\:{arrange}\:{the}\:\mathrm{4}\:{women}.\:{each}\:{of} \\ $$$${them}\:{can}\:{take}\:{one}\:{of}\:{the}\:\mathrm{8}\:{places}\:“\_''. \\ $$$${there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{4}!={P}_{\mathrm{4}} ^{\mathrm{8}} \:{ways}\:{to}\:{do}\:{this}. \\ $$$${so}\:{the}\:{answer}\:{is}\:\mathrm{7}!×{P}_{\mathrm{4}} ^{\mathrm{8}} =\mathrm{8}\:\mathrm{467}\:\mathrm{200}. \\ $$

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