![[;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_ukvLI8udi3nVynIrMKagDE6bS4PDNw4dAmglLGnuxBh7TOM0FMhaFQfMcDSgpKdmvnGPiAMMrSDtNX5Ryro_rTVN2ScWSRuR8QXD6nVltufrC3Y6txgP3XODJ6hftIhsk4l8hgJIlX7tYFzZsSj5MWZ-47chrYZQ3-Mb66paWSVoWWNFCmAymq2C9OzwTHUfXEhZofZrVLdO9scx2Y1cqQHsWauCk8mpVOQSuCiZvdhGDdEIHBHyT67XCZzEEM-LkgrjO0MO-MmEp4EVSRZBOFBaj2VK1fGENtHtgk4MBvKC-gLIJVMulPbnGCNQEAYuLn2Q=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_vUaagFS5-L4vqvzZBYd--1opdlVhVsPjHk19bKEw2A0dAOkW-_4ca3aOthusAZh-QkAZ1lMDREX9BFz9NgmqNZ_zAVVNvVpiTwJxkyAXVG9OpO8atNNVsAME2gZ58e1Lk=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_s36gd9x7YuYsLO-s9xVrzQfxbIskst8hXLPHongTwMaDGPjSfIVKpdH_RTWTS4ru67gtGY2_m2C92w3efY_s6zHmesGv63OSw3FRtmX1Z2fJu1lp1hWJPxsoyfRbQtEIiAbyk07F-GXNGHAAZ_kvmYsIFxwB0M0vAnO2-lZYxBJtZgqgIp2ptYfG0A7HUdacNbeWKobLY=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_taiLaYA3hEstL84r9yMeONeXe-axX-H4I2HO_UWyZWjQRoXtKHfpU0xVC_2TtFV8rZ5hCOM7K2_khl9YiTQgDdpoggPj6rqcwzRnm6kXL7_UKu3RvYlXnNlQweQqcVVaF4AeYGf2KzEn3mlP6t6lhCSSndf8wvdMAl54BhdxmqbQ=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_t22GmKMSHyYBJbEe2lxkkP1DW29O52JfF6VjFs-cxpR55SXsrNc5I9_LydZaaS9j9yELE1kQvX5m_Mw_aNHYzX_2ZjLVlYM_nUpcAyDw=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tVRhZe9Ma0LzaNqceZUJJEfpDyyXQNuy2J7Oxdr7S2osnFoREvHbtXgJNzOJGKdDschUC5hdmCo4WBv8NPOfj2XZbocBE2mwx9J61W5MgMiDjiuiMB2_-W7-gMJkzMqqsxRtU=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_tGyrlhmtPJ7P7ByFLSBsMZ2tcDZOaoj1zXh2g7nYvmoTI8nHeESTlzJdOL5R0ptOeuRo-Ay1ICev3aISlT=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_txJH_9iJFUuhFdIO5l-e9btjPFmBNzsYK-BYaiHABmHPLCPIwA6gNNPEfvN9S0GOBuqA0m9Hk-pFb2WYzQpc94b1iS8KCEtobe-0ptww=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_vnhKM1hhPbQ-vDt1wV4uQGaCzFsf_WjyjXtFL2--0yS2IHqZRVTUrUVS0nBKijaFy3cVzdiyfS_stGAwdRoCBCfaNYl_MsvhD_PKrEEEohfSGNSNaBMu4JyMdnfB_9ujGjq85xg-1Z-1plRRpYux3V9TnY67QRgi1Yh5yV5YsINmA-XEXPDi5d6K-MTUPaaf4-78x0N_gA_KrL-XFY02ZmBawC0xHeifCkQ1fRuDyrN2Q51Ajw=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_sR0z8BEWCpR1Mrg81xUYB-YWHLpa9kbAC1Cj3_B3Zsgi1A1-trjqw9TF_xzmkvszebU4_C9hFeTeR7KoA9vYOtZACCGDVckVmjGDVOph5XvoR_S4EZt6qZQS2__oHgAUjycyzQBTFkqVm9-hPkiZurRKfFiLnve4CtPDpudyFENKk1Xqrfr9bXF_2N7PhoH7Pte1nZxPvFaLylL1Uo_uJwOJP4MG7yNtd1fjng5jilQQgpMDY=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_t3fTvLYZFPoZCs51GQ6wWrGYzLKs4tNwB2qTK5pROqds5SbkCytdiCntsLWPPznpZZHasnvix8Lar3PMspYe7rrXAr_kBYlWXVyTnK9123PLkR6BaOWsT2=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_vYb0VB6U4eqXlvWIIauBFgTBJlKcCBmOuEuXChvO4PAVwauA1CJsadVPmT8g_WVtsqcJWgq947iapzOB6myB1FgVt62H5lUx52D7V9cVKF2jx0_sCpSEWadlfixOrKLnJCwbIaogklqwe5F-Q9sbl_jkTT2Ifk5c9t682K3nfoiSCqjvbqwup2TLZ8KWujN0A0ZY3xOL2EE3dQYg1cZ0NA4sappPWXmToaMWMrHMMKoprUiG5KKislM-hdwMucXmtsTgxBQH4YNcMaWsW7YOfy3zpEa4zPK3sm79sX=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_vuv3nu5aA9Dc0G35TspeLnceyIylD875fT8OAtY27AGTkwIgPdCvkH82b5gwbUtREnsjDjX9f3FmPw4yexh7Dn9-u7ZdZA_R4QCWjEFbUg04IyAax7HwMzpWT-7NrfhLm3uu6VCIIxcIpLI4AK0sMhPE8jkyuH-Lo7CAu7naeO7sEZ5lZ25a-6Fpu2x_H28b2AINOaPsOLBZDHb7wXdc-zNPG7Em0RyW9-Bw=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_vWc8d69bTSnt8KccaTjTBxtFD8W-HD9dKGx9T7O9Fm9HinVrOLygmSqftt9fCmG2CPdr47iy1_8eSwH8aenLId-pNfHvG08aJTJZ9C5vq2gGd53j48NAfwXuIfDfxxtJ6igVJjUGovEafzxSU9zlnowgtcQUhyDjqdUp5dmXQpgpQD5x2dcoLG0WPf92seaZBZOPbdJuUhHMk6HROB7LQsSfX9j1MIThIka0ZkbMWE3QTsrOtJBONmntGC2gvY0-UIAHbco-5G1SGfFepMUbtxdDVs5HzsGcxQ3GIR5n9biP555NJBxsa8iw7D4qCunnfp=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_uAsNsGG-Jh4c7YMSQjW8lkU6UKMNEYAqOz8N7otDvciRnG8v0lycx5ppGSw9r27A65L33FFQSm-wmB9V6bz_JQGnVbr7bOSKUsYJz96BLRmoExj1wJF4cS9imJO1h6NJWn=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_u2qZAcIDrUt7Ookr6hm3Di2NziVdufOMjinBfMLkcmj6jgTcsqXnHJ9o_SP6o5oY0BvJTd_b9prkW7ZCWYrUe7shmwf0XDbaBfgfCcGo8KNQrlfZ3iKRRFY9Ke5WqQRZf8dw3ufVIanx5lfKPtqMGCz9ACWfRFajDfNL694r7ocLt11A=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?!