![[;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_v0z4yIe431P6Nx8Au6eOpi9JIiLUypaZllCKgAmB513Xe0g2uVjCOXaNGQdPU1tnGZGsZ0V7_-x8l8xR6-BC_i6dwjQpMlZI8LY4w1DTEfwTxoJu73MZRPqOb_as7QAFwxEOdmfUOfFlkkuLE9hf9_N0ryicXmixyKEGhUUAlhvsCTX_BIcBk-aECuIvV28PF4A4I9yL3MuWkXgFm3eyXYV2JBTi0815rZJsCKN73QSjYxM7unOXOpG0GmYnGifTNk4bmfE2IYvCPUZSyuW3ydURykAiG-h-0AI7U8S4etfOnrgU2exZEgsbv9AVC260zUNw=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_v3CKgBqd1aR9Ad2_lY-Rjs_FbXv7lIuuWnzTtxAgPY8PE8LH2IVIyjnZKEwZNK1LggmRMYCEZivXMLoaKO6TntJU4WJSDFtXC0ozTC2qIobujfKAcwMRz-ChlAtP-v5Ac=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_vRJviK1jXZVDoTuZC49-g4RF1akozj9m4JhUxo7wtfmKfl-rz4zaI8lnKtFXdkB7LVuJntE87ij8i-z3PRC6ud3wA3OHkHiGA9-YdPk_sGIGFN5Fu0h7UpKKjYYW3Rr6wilwBHgXMNPAMRLRsVOYYHV7FnActZYPolm14GTr20Ale5LSdfn5u2_L0sR2vT0oXowyrVRa4=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_utmPplYBa541q8jygV_3mZXwmr6HBNfl9tTDVPbT__5yVjWH2ZQH5xZh0GMy2kqkY2_m7seoMS0ufW-C3n46tyfl4Mjwrv68Qm6EoskDA8w3VUJyLYT2uXEAWxFM6KUNyNr5oTAiwH0yu9bmpvMibBEL-u-NLhM6b6F0zLNsizew=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_uanaW45t-Pz6ynl4dFF6lJtSrrTVUJRd8m98gvS23Uy1ZyFzEFPvNBG_g0HoOSU3omHc9fe8BBw5q9tEBZgYd_680rVr7AC40IlSfkdA=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s4RL7ouibPz0zaRge6F13rIXuW_ku3Wl7RCNGbOsbj1IC9n2CIfHbR9DzgaKS0Gj-Zzprvo-YKdLh3XRlb853KyodfB4nb9jlkJkaWERVqGbkDmmiOU8-SRTkJTbTSq0Wdd2o=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_sBu2dmwmbLSi-8-uhChmBbjy7Nnkx_SWvccMAKGGS4hHU2WXh-ggIorJeohg8FBXcVI5Zg2-XvZ4e_Af_N=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sQwPzQA_iUKxgJE6UdJ6gpsmnoA2V_ViZz1rLFnl4H1FJKayGUOlX4ii_vG7MtJoB3jq64S3tpdR-ygqdf5vEwZYfPu6QiDSqn9K7kLA=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_tOwotWF4IYdZCW1UsvYZTEY6Lj8mrv4IsNLNZItEvIjSszdO6FqO6s-1S3WtIccYMun8wfdnKPulFn_0XQOkOiwqdobqF2pnD1LpCYaXbMzCTvGf3cHbR6wT4o-RTiZkFSr8xV5rsPLV5DGVVU9lMG1Ib3KdUB8DqE9GJaIdkg6cAvwp8Xfizaz6-PIWdAFVB6a5-JbuhtxXzTaY9tEzfWgvbhT6yPDQqPgF6lzE-RoiGe9eMw=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_t9ZbJK-fMprVcp_BCKgiMV8dMkBBwu8SVb1pcoPl_4EKFA59DYmjinx84dCHIVSXXspfiVn5al2PcucC9t2N6TbatDCZxPI6euw4jZOI3xAJEtUzNxQ3OaFG3lsBbflyHFQI4wdMalSg7Fzpn9EpPuuIVwbeN_rr_XADf5hODlJkepTxCi_MrSpF8TwW-U5juATeyFE-3EIOE0Zi50lu97vd87YtLS2alKoy0oglwZXKgipSM=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_tm9dKQ99khYM7x6v9LuwBIr9bAZWgcA4f-OtLWYS4lF3uLvvRpy4iYTXDj3nEYL9i22913vjxMUA0nL7TDfCkMWVig0id3sQMHvDWFIsJ8nVtwZHYpfGWb=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_uwZW1ThMUWsK8JZ06CA4d7omSkcalMqLygc4Yn_wZhnTBjjGjGem5kNhv8qJtLI5UgNAOcb1gDG1a2FoiJrjCnAE4Guhcm2l8-xAegqbMAZA38elp1whU2sp-1EMLDCI6o35k2Owxdva_tgMS-JQYISR_ULF7KJ-LrdOn-Z9MfhMq98Dg9pypDX_Jju2rxQgOLBEa_lo7NhGk-qR6TrsLbqdH1WnTXTIMQUaK37jF1oKTmbdFjcaIF1qyhB7Cz3fBdPEpVhXlY3DBWzLIuhXtujAU9bGYm41jIXYgq=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_sjl7jHiu4oQl2iC4U1r4eGo9mvNKWEgGw369JkW177lxqhPFgbRJ3sWwfUT8Q5b2WzxPAbZqvl4RnL2o7P0j71aOD9PyOl6sEqARByj0lFg5Ef433diIFhydOIa3A-g3XChwIM9X4jGKelInzezqo1sVAAXG0j0ntRPBKfz_7ES9GcfxTZLn1uptMicnFgiHkCYCwLbNwF-Knsbsrycxj1TIbn0Wwe7sHQTw=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_swCWo8uXBEMlcDfsGp80DqWhxbUVgscdepbhlKAoIX2h6N_Ur5-3QpiGcP0wAh4PAvRoBXCptOFkd2JTFQ5q-xxnIEbrImEU4J4XOgLzNJ0i06WBeiLawrWyRTjSS0771z9RiI4RulmaDi6xrmkLO4qOVX1pT10_efww29J5X8b9WGOqsPgdmaH8w2DHyX1qmqo59_R_gIg-epxV6Wpdl87652yySaYv75m6qFxRm5K7CJRRnOWpw0q3bojvMiWuUhVqqP9oaIOTfwNYNZ0_n1HZg2Axy2RHHI0Pws3HUzRZ_ujIZZyrP6rOhy45Fv7KVL=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_vfFgy-jg3V6uvkRY61GdFwg-ftf_9Xo-1eMNTCCpAYgFuS7yz_jLW67T_25N2VDtUzLPPDtJhzk0KLPMbNOE7w0MPdRWHjNKfeRxs635zulUowwb7MSXqf7x8wK0RerE_Q=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_u_WI74Oy4QZO7UUXBslVhoN4IOZbQ3p_FsWMlr8WqA2QnJQR55xX6vJj_eELdKFYmlian86qTrX0TnbNYzPJjfINdfI4jx4WA9H9t8o0bp6oVAZ0Vt8H10U1X8fjbZNvVrU00bGmc_fbkLS18uq8cy1gALstppoBtZ6_t7Z_hWb4y4Sg=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?!