Question-145290

Question Number 145290 by ArielVyny last updated on 04/Jul/21

Answered by ArielVyny last updated on 04/Jul/21

$${exercice}\:\mathrm{2}\:{Nombre}\:{de}\:{Bell} \\ $$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

1) L′ensemble {1} a une partition : {{1}} L′ensemble {1,2} a 2 partitions : {{1},{2}} {{1,2}} L′ensemble {1,2,3} a 5 partitions : {{1},{2},{3}} {{1,2},{3}} {{1,3},{2}} {{1},{2,3}} {{1,2,3}} π_1 = 1, π_2 = 2, π_3 = 5 2) p∈{1,...,n} et Card(A_1 ) = p Pour former A_1 , il s′agit de choisir p elements distincts parmi les n de E. Cela se fait de C_n ^p facons. 3) Il decoule de la question precedente : π_n = Σ_(k=0) ^(n−1) C_(n−1) ^k π_k π_n = Σ_(k=0) ^(n−1) C_(n−1) ^(n−k−1) π_k et en changeant d′indice p = n−k π_n = Σ_(p=1) ^n C_(n−1) ^(p−1) π_(n−p) 4) π_4 = C_3 ^0 π_3 +C_3 ^1 π_2 +C_3 ^2 π_1 +C_3 ^3 π_0 π_4 = (1×5)+(3×2)+(3×1)+(1×1) π_4 = 5+6+3+1 = 15 π_5 = C_4 ^0 π_4 +C_4 ^1 π_3 +C_4 ^2 π_2 +C_4 ^3 π_1 +C_4 ^4 π_0 π_5 = (1×15)+(4×5)+(6×2)+(4×1)+(1×1) π_5 = 15+20+12+4+1 π_5 = 52

$$\left.\mathrm{1}\right) \\ $$$$\mathrm{L}'\mathrm{ensemble}\:\left\{\mathrm{1}\right\}\:\mathrm{a}\:\mathrm{une}\:\mathrm{partition}\:: \\ $$$$\left\{\left\{\mathrm{1}\right\}\right\} \\ $$$$\mathrm{L}'\mathrm{ensemble}\:\left\{\mathrm{1},\mathrm{2}\right\}\:\mathrm{a}\:\mathrm{2}\:\mathrm{partitions}\:: \\ $$$$\left\{\left\{\mathrm{1}\right\},\left\{\mathrm{2}\right\}\right\} \\ $$$$\:\left\{\left\{\mathrm{1},\mathrm{2}\right\}\right\} \\ $$$$\mathrm{L}'\mathrm{ensemble}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\}\:\mathrm{a}\:\mathrm{5}\:\mathrm{partitions}\:: \\ $$$$\left\{\left\{\mathrm{1}\right\},\left\{\mathrm{2}\right\},\left\{\mathrm{3}\right\}\right\} \\ $$$$\left\{\left\{\mathrm{1},\mathrm{2}\right\},\left\{\mathrm{3}\right\}\right\} \\ $$$$\left\{\left\{\mathrm{1},\mathrm{3}\right\},\left\{\mathrm{2}\right\}\right\} \\ $$$$\left\{\left\{\mathrm{1}\right\},\left\{\mathrm{2},\mathrm{3}\right\}\right\} \\ $$$$\left\{\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\}\right\} \\ $$$$\pi_{\mathrm{1}} \:=\:\mathrm{1},\:\pi_{\mathrm{2}} \:=\:\mathrm{2},\:\pi_{\mathrm{3}} \:=\:\mathrm{5} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$${p}\in\left\{\mathrm{1},…,{n}\right\}\:\mathrm{et}\:\mathrm{Card}\left(\mathrm{A}_{\mathrm{1}} \right)\:=\:{p} \\ $$$$\mathrm{Pour}\:\mathrm{former}\:\mathrm{A}_{\mathrm{1}} ,\:\mathrm{il}\:\mathrm{s}'\mathrm{agit}\:\mathrm{de}\:\mathrm{choisir} \\ $$$${p}\:\mathrm{elements}\:\mathrm{distincts}\:\mathrm{parmi}\:\mathrm{les}\:{n}\:\mathrm{de}\:\mathrm{E}. \\ $$$$\mathrm{Cela}\:\mathrm{se}\:\mathrm{fait}\:\mathrm{de}\:\mathrm{C}_{{n}} ^{{p}} \:\mathrm{facons}. \\ $$$$ \\ $$$$\left.\mathrm{3}\right) \\ $$$$\mathrm{Il}\:\mathrm{decoule}\:\mathrm{de}\:\mathrm{la}\:\mathrm{question}\:\mathrm{precedente}\::\: \\ $$$$\pi_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{C}_{{n}−\mathrm{1}} ^{{k}} \pi_{{k}} \\ $$$$\pi_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{C}_{{n}−\mathrm{1}} ^{{n}−{k}−\mathrm{1}} \pi_{{k}} \\ $$$$\mathrm{et}\:\mathrm{en}\:\mathrm{changeant}\:\mathrm{d}'\mathrm{indice}\:{p}\:=\:{n}−{k} \\ $$$$\pi_{{n}} \:=\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{C}_{{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \pi_{{n}−{p}} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\pi_{\mathrm{4}} \:=\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{0}} \pi_{\mathrm{3}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{1}} \pi_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \pi_{\mathrm{1}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{3}} \pi_{\mathrm{0}} \\ $$$$\pi_{\mathrm{4}} \:=\:\left(\mathrm{1}×\mathrm{5}\right)+\left(\mathrm{3}×\mathrm{2}\right)+\left(\mathrm{3}×\mathrm{1}\right)+\left(\mathrm{1}×\mathrm{1}\right) \\ $$$$\pi_{\mathrm{4}} \:=\:\mathrm{5}+\mathrm{6}+\mathrm{3}+\mathrm{1}\:=\:\mathrm{15} \\ $$$$ \\ $$$$\pi_{\mathrm{5}} \:=\:\mathrm{C}_{\mathrm{4}} ^{\mathrm{0}} \pi_{\mathrm{4}} +\mathrm{C}_{\mathrm{4}} ^{\mathrm{1}} \pi_{\mathrm{3}} +\mathrm{C}_{\mathrm{4}} ^{\mathrm{2}} \pi_{\mathrm{2}} +\mathrm{C}_{\mathrm{4}} ^{\mathrm{3}} \pi_{\mathrm{1}} +\mathrm{C}_{\mathrm{4}} ^{\mathrm{4}} \pi_{\mathrm{0}} \\ $$$$\pi_{\mathrm{5}} \:=\:\left(\mathrm{1}×\mathrm{15}\right)+\left(\mathrm{4}×\mathrm{5}\right)+\left(\mathrm{6}×\mathrm{2}\right)+\left(\mathrm{4}×\mathrm{1}\right)+\left(\mathrm{1}×\mathrm{1}\right) \\ $$$$\pi_{\mathrm{5}} \:=\:\mathrm{15}+\mathrm{20}+\mathrm{12}+\mathrm{4}+\mathrm{1} \\ $$$$\pi_{\mathrm{5}} \:=\:\mathrm{52} \\ $$

Commented by ArielVyny last updated on 04/Jul/21

$${thank}\:{sir} \\ $$

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