![[;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_sFkWsQdWPP8t-wtr_1VY0EUsnA7wSgZ8jZPygQVEMdPPM_2xPi8VDJE6vK-LhBYyUFkphs59pS1lurA1zpgYA9WSwN6dQh0ZVL_zVJ8rbm9hC3KHNjajQ4MhjeKBQaORQAL6Ee3AvfifT80_UVjOe3AhqslyGJzZfkzRcSoh5vvS1iMBFrxNlB1NuPsMgkvdCxnnM4NHYLW41pRM4jtkr-GNq7wbkc9dtjfYUMdjtvVi3Clfb7RIRSkGlW29LknA54lJ_XNPhI2CsmC7gO0LOSnQB0Nc3eRT7MmWFAqclPgxbANdpjn6G88Ya6Qm-EN-HvpA=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_shNru3KScdQ3nwyxABXzrZI5SOFrWHSBPx_cH34YnwmEmYJkex_pE0AUn1QrNkEado1Zun62wesn3FqpYOE2PmMak58ht_Ml8SxEVvAWaLKSSlIUza8TK5tGgUdr35oeY=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_vAXuDWVsrYMYC0e4s_OD-7L561pswk8-WTeLaMZRXZjYhcVfMZe4t6wfVILOSaaXSOf1-lDjzuS6S0QiqU9z5hi0Ba_enA7EjGOwruvS9MYqcSrr69FLwR5uj4yHnk_xDyGxfeE9krFmpz1jy_caZUdn6EwvnuqVb1ZZwHpWrtcB9FyoD2WjfoR5LHiZOMIqw1-48SJ1w=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_t6rEzKwYo42i5Suv9_NYlu1GsP0h0nQAGqVloRfulGshsAEIeMmRZfkblyE_ug3fq6GR91034m4u1gtiySvwVte0YkuXoy25sPxzcBkze1PmdmTHxMclvyELqBb9fwBGTiVNItk7ARtp7nsNioapJWJuJ6gTSDIs2xZ10KlUEufA=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_sM3aPxZS4L-Xw06CvlXwhJZJ5KFNVmsuCBfc9BrR9PhCpbSEbM5yoCm7VmNYNuqKpcQQwOvBAebG-UsJL1Vrd8x5tflgkU_JXTRaiVVA=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ugGujQxDwbxlwG9xWyg0L2Q_PDYmC9GVeXX-FpKQxumoeRXE5GQMh8d4nqEoCk5K9e4W4in6PAZU0CfV2s9RU7dON5lrypa6Hi_qrigCr3TxxDu8IG9qEy5U2rCTcy0kH1htU=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_uzsfDZFlxQCsmri2XcSQLDx-ngnoXMN8IBswsqN_ZN6PE6UPSJk3vynMlsYQ-_zy8QECyKUfWKbiiCFovB=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uFI9OJmE9Hx6g9X6MwKHCcRVcRhnq9HpnCd3JAWfMFgtTNNmSsScZv1dJXzYsNRlJ2bKysSS02GRwpn8exrRoMpn6pWmqK3C_wOLi6AA=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_t4RE3kmoPa4IQ95Hb4k1ZDWqlVixVtf-fTKX36OMJ37AwwxDV3-58viBZeSe10Pph9zeSpK-upjda4P7uIM1TINZu-YpQKUJ0BWLhOIjRCPctdZ2r8Da1VPzb2HOrynDOrbhbA4W4wowqcg2BGg-rj_xrfgi_WtNP0vSxkH-WC2xm7SCVzPSTXY_ho6xSch9m_DXs56HB7Is_62_yjauWJ73R1mvIIFqslupwI2wjz_Es35TiQ=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_v8u9tEh9WtyWbHBuV4hlFk3o0yYF6a4L-6W1J4DtIMBvJrGQ8hfMtCf0GXBvjS6KnPXw_76IYu2Pq4etuZLS0pCWDCQnDqv4mTgsDCbyDoWx4lHuwdYVg7ngcJ9MT2QWwqf-HF_dJS7-IER-wNYTIEeXFXJlsWNSk1dgDlGoyGC32hcIQ5i6ooD04fNg1X4Z-_LNN_8mBaH4ZeuWAnKaphZUBPfoQNMZyJ87S40Dn7T6dWRCU=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_sRQlNFUYrSYScsSflELs0J6xKEZ9qnOd33GzEHDoJ00TwT5qpaygvOwX-6zXcMp1qE_0COzo2IVyGUxfPld7eOU6RNXeq33QhbNIe_Fq7Syz-nc4SW3NlS=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_s97s7Lm2jmVHaj7O8YkLj1HR_ek2I4j1S_8AUrLl8XSXmFndmJD_Ut0hwG8pafYoXbwBavJTq6u9gaYUgCRl2CpbFFMVz16_pZE-c3mma9kwT0fBaR5XG9RVSDeyWJT0V4HRrAV4qBuYwHxFdF7_MwzGLWFC5s3mJ2cK_e2lLj52mmJKQnV7-3tIFG4VKG2qJA_9n-zpi8NOqZkZJdKbEn2lo8OGDrhWq-xIVGIBBcmCu_UjS3wI5gWkZRUEoHEwSXJWG61Jlv3m_NizNsLGHx-RFz4eVMK7QpkHWs=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_tgdUGOrgD2-rkk-KtIyiK0wtBOuSMm1PvNET0SiujsjwrNfCBfU03sD1pmzeapItWggDUwR04FKjI1T445xmRu09G4Pwz0n_ejBrk6tJ7bfPv9cq2WTcG8upjTl4tX8zFMiY9XfX7L-B1AcNeE-NnWqPyb2-KF32ascv2F_RMtXWy3WoSYMHhokL4VqMNh9s6rYd3xDFTqfGzeyQxPzjhQ1evNM2DVRfB3TA=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_uOSOQyZL_Y1NYepPMqtEQOUv-34s5Xqa5lU8CCyOzHAl8Z6wTUYsHBGttG1SIWtPMhU0eiXmI-TocpQTtbpS22P0Ki58NEN7vbM1wmkLORCjbxORrsTYufJskFXjwepTnNWII3A29jeQoQcC3kpdQnOwOnvr0F252TZg6-HNgHoN6mmsJrZRTiPCePkzwgl6VLz4h_glLKuNHBkbqBNixRzXbfX7QKgTgZKWNo11uClczU1XxQ6KGF1zkr91dim4PTsl6zskaQXSjAdPR4V0avN3V7nmt4fHf2rclgP6o2ekbgA7EZ6xp5U5YxIKdoLJjV=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_sQ2w4a5Xwhx77sjj2af1TwL0LAqnrJYcGNtbLrIFfuCsr92T0Xgl47gNgPvWMJP6I6k6F4hbTazxJMuAJEK3muXveayiN5td9IpHaVuGwLWZpO-1iKJD5nn6tyIZbqCki9=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_t8WjsM3P4CbW4igjiZckv9DbdD5JIcYAt2VJaSJHsKKHHKmXC5OFEIdN5CHNd4RIinEJ7j_75TroN7ncAOGEjbD6ddw1vAyNbNGkzrDViuMDhI_Y3A4WZ8kpFZ2_nDbP6ppTHXB6TXtxbK95f09gM34q3_Tfw9Dl2p3d7qICMdLryBQQ=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?!