Math Question and Answers


Question and Answers Forum

All Questions Topic List
Probability and Statistics Questions
Previous in All Question Next in All Question
Previous in Probability and Statistics Next in Probability and Statistics
Question Number 145290 by ArielVyny last updated on 04/Jul/21

	Answered by ArielVyny last updated on 04/Jul/21

	$e x e r c i c e 2 N o m b r e d e B e l l$
	Answered by Olaf_Thorendsen last updated on 04/Jul/21
	$1) L′ensemble {1} a une partition : {{1}} L′ensemble {1,2} a 2 partitions : {{1},{2}} {{1,2}} L′ensemble {1,2,3} a 5 partitions : {{1},{2},{3}} {{1,2},{3}} {{1,3},{2}} {{1},{2,3}} {{1,2,3}} π_1 = 1, π_2 = 2, π_3 = 5 2) p∈{1,...,n} et Card(A_1 ) = p Pour former A_1 , il s′agit de choisir p elements distincts parmi les n de E. Cela se fait de C_n ^p facons. 3) Il decoule de la question precedente : π_n = Σ_(k=0) ^(n−1) C_(n−1) ^k π_k π_n = Σ_(k=0) ^(n−1) C_(n−1) ^(n−k−1) π_k et en changeant d′indice p = n−k π_n = Σ_(p=1) ^n C_(n−1) ^(p−1) π_(n−p) 4) π_4 = C_3 ^0 π_3 +C_3 ^1 π_2 +C_3 ^2 π_1 +C_3 ^3 π_0 π_4 = (1×5)+(3×2)+(3×1)+(1×1) π_4 = 5+6+3+1 = 15 π_5 = C_4 ^0 π_4 +C_4 ^1 π_3 +C_4 ^2 π_2 +C_4 ^3 π_1 +C_4 ^4 π_0 π_5 = (1×15)+(4×5)+(6×2)+(4×1)+(1×1) π_5 = 15+20+12+4+1 π_5 = 52$
	$1)$ $L^{'} ensemble {1} a une partition :$ ${{1}}$ $L^{'} ensemble {1, 2} a 2 partitions :$ ${{1}, {2}}$ ${{1, 2}}$ $L^{'} ensemble {1, 2, 3} a 5 partitions :$ ${{1}, {2}, {3}}$ ${{1, 2}, {3}}$ ${{1, 3}, {2}}$ ${{1}, {2, 3}}$ ${{1, 2, 3}}$ $π_{1} = 1, π_{2} = 2, π_{3} = 5$ $2)$ $p \in {1, . . ., n} et Card (A_{1}) = p$ $Pour former A_{1}, il s^{'} agit de choisir$ $p elements distincts parmi les n de E .$ $Cela se fait de C_{n}^{p} facons .$ $3)$ $Il decoule de la question precedente :$ $π_{n} = \underset{k = 0}{\sum^{n - 1}} C_{n - 1}^{k} π_{k}$ $π_{n} = \underset{k = 0}{\sum^{n - 1}} C_{n - 1}^{n - k - 1} π_{k}$ $et en changeant d^{'} indice p = n - k$ $π_{n} = \underset{p = 1}{\sum^{n}} C_{n - 1}^{p - 1} π_{n - p}$ $4)$ $π_{4} = C_{3}^{0} π_{3} + C_{3}^{1} π_{2} + C_{3}^{2} π_{1} + C_{3}^{3} π_{0}$ $π_{4} = (1 \times 5) + (3 \times 2) + (3 \times 1) + (1 \times 1)$ $π_{4} = 5 + 6 + 3 + 1 = 15$ $π_{5} = C_{4}^{0} π_{4} + C_{4}^{1} π_{3} + C_{4}^{2} π_{2} + C_{4}^{3} π_{1} + C_{4}^{4} π_{0}$ $π_{5} = (1 \times 15) + (4 \times 5) + (6 \times 2) + (4 \times 1) + (1 \times 1)$ $π_{5} = 15 + 20 + 12 + 4 + 1$ $π_{5} = 52$
	Commented by ArielVyny last updated on 04/Jul/21

	$t h a n k s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum