![[;C^{0}_{m}C^{p}_{h}+C^{1}_{m}C^{p-1}_{h}+C^{2}_{m}C^{p-2}_{h}+...+C^{p}_{m}C^{0}_{h}=C^{p}_{m+h} ;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vrpYlPMBBKnUqoWpst23Fe8ez6SdiTzvplQIHjLjRcjBnIrg-hoehq4eLcQlWn8o7QHprCdQxXawDP9KrLAX1QgMD6TwUT82SnFor1LCu5RVweWv9nH2B-2An3d-LIsfW5cwcluwX8OkAOavHiWLdplfeCKz9e0xgeCEEvlqs88CNOZ8NZ8GqnYFk0mF3UH5k_Gm5fF5QNN0NGxanSisfotm6E-qkgo0y_AS-hdI83oYAKhcL1WG11wvVnW48RgkCFU4LfJxd7prgpTalBXPJpBjvXxlaaeEPOEaD7WUu1GWcJoNr4m8CJe0RBoGK82WAkHw=s0-d "C^{0}_{m}C^{p}_{h}+C^{1}_{m}C^{p-1}_{h}+C^{2}_{m}C^{p-2}_{h}+...+C^{p}_{m}C^{0}_{h}=C^{p}_{m+h}")
Vamos provar essa bela identidade utilizando apenas argumentos combinatórios.
Quantos subconjuntos com m mulheres e h homens temos em um grupo de p pessoas? Podemos pensar de duas formas diferentes: somando cada subconjunto possível ou simplesmente escolhendo p pessoas das m+h totais.
De fato, o número de subgrupos nos quais há exatamente k mulheres é: ![[;C^{k}_{m}C^{p-k}_{h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpejv2Gnd-DBzaIyiOrSmiDdKwgrqlG9BiGpjqs8Zg6k34OuUv-PsJegrGRxNAnZX-wvQ_iO0G3ZkXS_maV3-D0QHx8guoA94BO0g2zuhe_tLqzpw-3AQj-0jUadOhvTg=s0-d "C^{k}_{m}C^{p-k}_{h}")
. Logo:
. Logo:![[;\sum^{p}_{k=0}{C^{k}_{m}C^{p-k}_{h}}=C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uxJtyj7qHySfIZIva2X1Wu-I18JNGoa5DcG2EgHho_4FQCnESAo3IeZY3kwkXjXVZhUumGx0G6bKsmVf-Vwyn6kzuudd5smWmHXQcgLM64OP2FcXgxzSYXDTloyA60f1pihNLjAzUuGXY4XVRGP7WDihBL8wWNkyiMevRk7iFlDe-RSW5tVQJWXgXkSu9VUCietLogA5g=s0-d "\sum^{p}_{k=0}{C^{k}_{m}C^{p-k}_{h}}=C^{p}_{m+h}")
É evidente que poderíamos fazer uma demonstração simples aplicando o Princípio da Indução Finita ou a seguinte identidade:
![[;(1+x)^{m}\ .\ (1+x)^{h}=(1+x)^{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2F6j6iAaTiTY3UQoBnDEx0304vqNK43v7vXaTZ6Hqj6c3Mhl_YFJyo76Pn0CmPhdXSsOnVnEhwdsgZzLBrvTr9LcyQnhjHu9WEbNmRd_1Z5fAtCfZ4LC5CmfVb2QplKgb4feT2_2RT75EuECcRBGb0ZkeYOoUN59uQ6DmD3QBHg=s0-d "(1+x)^{m}\ .\ (1+x)^{h}=(1+x)^{m+h}")
O termo genérico do desenvolvimento de ![[;(1+x)^{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uQ8DAjcRrk5Hl-A8V_skWmF2dAlCAupQBIW0jQwfEAKvhAGcy5TnOPIMDmG3waB6F4hicLAgfqM0XsoAgQL3zD_zPun1QcxyCJaxVa-A=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLkgi_jz6mFi3AYydvhtCRLMPzFXprGdKjD4ekkxdCdwM0xlIsXxwTam0q7GocJBu1aldkjDO6kGJfbdLgArDJGQQxXaaTCymJXkxBMviCfy2Ihjtlw5u7RIOdFFg8Ju1O6SI=s0-d "T_{p+1}=C^{p}_{m+h}x^{p}")
, o que implica em o coeficiente em ![[;x^p;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ub-tNRPtW461GUyxFtQ8fuUh0v9KTlg8oXgYlenuIMfXa5WlRGCxgg63VUTKnoUDUCBJ6Xn6eYPTm36_VL=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vY0FwdUEmT_0Tx4UbVgoO1cNxlfCuGI4J5O9Yham2_Fdc58hCAAxw02BktkmTTXILmK_ZtkaZY-rZzZoeKMRNZtFNXGVGRNtD1Bb0BDg=s0-d "C^{p}_{m+h}")
. Por outro lado:
é , o que implica em o coeficiente em ser . Por outro lado:![[;(1+x)^{h}=C^{0}_{h}+C^{1}_{h}x^1+C^{2}_{h}x^2+...+C^{h}_{h}x^{h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sGRPvC2ynmuVygNQvegqqj5HW4Of-wGJVT_vSA1dwWFJYmijqfX4x1KvBUem2q752UTMs8c1p4LkjDvH8h_PZs6k2DTIVmU66lN8uvGTv-eVqjVwkN2AMTKoIj0S5237Obk5z60Ke1fTzGyG8NI6D9I7-ZjVLm8X0aG981Q-8nEtdKPyMH3UplaoNytj8UEHjI-ojFJQ9RZAlgedSPQA6mdii37XRYKPH66kAw5F8SlJ7eEI6_=s0-d "(1+x)^{h}=C^{0}_{h}+C^{1}_{h}x^1+C^{2}_{h}x^2+...+C^{h}_{h}x^{h}")
![[;(1+x)^{m}=C^{0}_{m}+C^{1}_{m}x^1+C^{2}_{m}x^2+...+C^{m}_{m}x^{m};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tgl0XkHawWKnYG__4LkkuHxtn2JayZPFej5WMPTXneSxV30uA05EYjyaNKv53jYNb11fPsFgoazSANAZkmT0ABa5r6WWfHFTiHxVNDSZ4945LVlEtdQFJO-WnqxcsfyTLr42cv5Tva8sNUpD9EVM2O_Z13p9ttuJzh5aj-BHdFRkSrkKFAKLfSudYWOH375PLkhMiCinNvC4_1Jtex6JnQD7p2YT_aE8PCXXn8SP340NVz1vw=s0-d "(1+x)^{m}=C^{0}_{m}+C^{1}_{m}x^1+C^{2}_{m}x^2+...+C^{m}_{m}x^{m}")
O coeficiente de no produto ![[;(1+x)^{h}(1+x)^m;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tEMjUCiu-qWSeUJrQv0n8TNHoXCd2cKbPLdbAGqp-A42fNoGADqwLopEc-VT3WCvf6Z74S0ndDsKsySp0_R5UTmitLCRQ7z8id1CNmJ9dNaGZkkZj04XHO=s0-d "(1+x)^{h}(1+x)^m")
é:
no produto é:![[;C^{0}_{m}C^{p}_{h}+C^{1}_{m}C^{p-1}_{h}+C^{2}_{m}C^{p-2}_{h}+...+C^{p}_{m}C^{0}_{h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tCvBiz2L5FQAgZqEcVkfYHO7J4q9sYDse-aV-p7OXMToZSWOm5Na-70kO8r6jFmIqu5jKriNwtDCi286sGSq-uLCq5mGYW2-DaWoYSvwA8aVjbTxbpR6ITtBrwDSSEOmGhxomOa45RevIPX0wXRoO2-89i1Ay8Tu-WpKotSUPMfSNyxPFsS67YhdTGurJGZNqXSm_tQsWcE2CqaWmUzo-gkdzuJJzeRHBHhwo2w4vzAA3Luqkoh5t0q_7r84Mah4evGNdkP2Syq9LO-y7qERsEue-ZQ5-MO3XreNFW=s0-d "C^{0}_{m}C^{p}_{h}+C^{1}_{m}C^{p-1}_{h}+C^{2}_{m}C^{p-2}_{h}+...+C^{p}_{m}C^{0}_{h}")
Logo:
Corolário (Fórmula de Lagrange) ![[;(C^{0}_{n})^2+(C^{1}_{n})^2+...+(C^{n}_{n})^2=C^{n}_{2n};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sJmnGAuyTpTsCDbBYtGMPdFo2gEYd08W1CMMhX6XpvXOubZzcIdBWp75qV_jArR1RPxbRGl2XWJqDTEIlTKFe-p6VK4lrwPGz6Zbhkhe1CNMPJQswoMvhrg5JJTtNZRz_7ggZuvpkWkbhTBQ_fPOBuBWu2HRe3ahpgmnlRlrT7pmJrfUPjNyGFxJ4TW8UKEADHmIvIuCvnzQdUTIURgcHrHWNL1x15Qdiccg=s0-d "(C^{0}_{n})^2+(C^{1}_{n})^2+...+(C^{n}_{n})^2=C^{n}_{2n}")
Demonstração: Basta aplicar m=h=p=n na fórmula de Euler:
![[;C^{0}_{n}C^{n}_{n}+C^{1}_{n}C^{n-1}_{n}+C^{2}_{n}C^{n-2}_{n}+...+C^{n}_{n}C^{n}_{n}=C^{n}_{2n} ;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tFZZqJgVK5u1WNO1JRgP1YS8YPfALTbtIJ3diQyNrOxsOlyeLxensTBM4a_E4FkEyiM5S-AQwipGENAdZ0xnaw5KbV9CvFcJxfHIk5rawed3QPLYhSAcoTpjx-7CQFzSlfh0NbauvAwyT47ST9LW5-qJagr8yaayUmvhjeK0eaL5GKXQvEe5b1nNwDOkzQaN5rz_TemWP1H87lHBSI3GMVlsId2ulSQG77DorJ2EXxmTdJbC6ghLSLB0Kl4nV2L6bJS601g80j4gYg3f7BZ4844esVJQDuRJcYuj86B5cSLPFsPhIIAaeeYML-QS1HmMFV=s0-d "C^{0}_{n}C^{n}_{n}+C^{1}_{n}C^{n-1}_{n}+C^{2}_{n}C^{n-2}_{n}+...+C^{n}_{n}C^{n}_{n}=C^{n}_{2n}")
Mas ![[;C^{k}_{n}=C^{n-k}_{n};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s0ADgDQVOSWaPrrtjaGj9FzTDXt8wQSXHnhQWuDwvouxbTIElJeFqC6IhYh5oOiCq_5hLlAu02wxGdy6-TEMN6KCwzYAHXzUquE2ss2-3E3O5uCBhbWknQrdnXsmHNn-rY=s0-d "C^{k}_{n}=C^{n-k}_{n}")
. Logo:
![[;\sum^{n}_{k=0}(C^{k}_{n})^2=C^{n}_{2n};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sVUH1qFufpvXEuTqkwS0hcwOWuC6AeFdr0MdMHENKDXvEw0UlADn8qW2kBDwga97jTsTJUzIaQFax_k8vwgDNWN7VIkGG1YI3r0_EI5zxoI05gRImcc_g6KEkwvrsssOOEP9213ApVO_vaeEMnoKEZ4TWA_jWy-gM8TMXp_8ajvBQ3uQ=s0-d "\sum^{n}_{k=0}(C^{k}_{n})^2=C^{n}_{2n}")
Referência bibliográfica:
MORGADO, A.C. Análise Combinatória e Probabilidade. 9a ed. Rio de Janeiro: SBM
. Logo:MORGADO, A.C. Análise Combinatória e Probabilidade. 9a ed. Rio de Janeiro: SBM
Olá pesoal do blog!
ResponderExcluirTentei visualizar os códigos em dois navegadores distintos: chrome e mozilla porém sem sucesso, sendo que no último uso o greasemonkey, não sei o que fazer, alguém pode me ajudar?!