![[;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_tEL7DGQI5gugZLD2ocoFpu-uYhmH7bePIMaDZ96XxrknOW2gHPmUpOlietyQVhXpR-0hSXEiJxog-Cs_K9y3dB2t_VMZXRdZrEMho-S6aDs8QJgtKDZNiDTobGlQt1y-4CFX-C8SoFKihdfiUQRJ2RPgOaTF7U4SjtrzJ3gnErPN5DTSmdQu7EOxgowPTNotwXhrZ4A5H1foddQRFTjVVSKLVrzijK3flsf3psM_NBjEzG3UcTIysOcixRprlvaxpxj3Q1QHZNMmbxnjsOnfPXNghcOg9Xy9sf3mAu_Ap-nA_qEnss00MGzcEIuLGfVpSajg=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_vLne7Vw012dwivh8RfSb7Yp3kDhgtZI7dx1eyC8J33aPgKHcmYPE1p5kZnOQUMppYZ3Yar0H2SalBH_vMoCPEiDq1hSGGgVYc5xsYJ-pNtSic3IJprdu9xcejrD1aaA4g=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_uMtGf0hcVRPOyaF0dFKgj5LFO8cYQzkOb4lVydJja-n9YsHqCL48KfJNERo2Zh2Rn9_zzy2Y5jfgaNoSyTdYB37xfIoVp23znHhytP8qyQ10RtHnDJAcWQd3FxfNUH30YfpMCihXdZPm-3l9L0_EdbBiLEgKWVIjhyXcga9wTSgid2MATzTW0FPvEtuPWkm5SBoITP5xA=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_tlqCmtxsA7aoqWRTXAYblAiRTZACvc8GGZA2sJpgkkOVRV9aSsN3toLuFsT5aLqC5_NJuQ2QEd8wlqsG1OaRaAMFipjIbMKwPt_9M6n4jMo9wxEWm-wMOHWoX9c9FRHMYhpOyYM-rR2Y6sBkUQps1Sw0iGrtnKp4PJgidh0GfjqQ=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_tK7IGy0_DUUHWelfYZgZN5Vxhc2pLQnd9uBsyIr_B_1YbqubuLV9yauuTXNsDqdRLKsKRDauxRDZGvAcTyMTKGElG8_SH3mF3hVRbbzA=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sgwcntZd1HWF1bNKuBvdYX63SHVL4wdGKUbVOcD3CUF7ulpUgNRHaKX1VCikQcIHa6z0tL6Fw9d5Bxb-2t3q-7vb67TUwV7DRFOUjdQJPsV_xhLtmBIc4laujyl2tXshbKTvE=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_uATBeOj0fwW8X0rij1cT47UU5dc7C6sX11Ml45RYQLj5z5TB65nLtOdE9tqXkvF1g2NXobg3dxHiAIsn-v=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uuKjHb1eqU6BSnprdIPaqf-ckIebCEzb6iWP3wuSe1avM7PQD_QrOs1GaO23rlr8APWG1wKWWwZZOJ-g970PCNg2BFHWMbdV7fDeFJ6g=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_spZtQQGLgSoRg2tc6rOQpf7fnnHFkXVBNVAOyLxq7jAGbpL8SVQCfXqC3rHBZjXSBVOsdqcercnnRWuyjro2RAfjf7DFsMNYniT1s5kcRgrQKIJa2bbIlgCz5VKmvte9YoAW_41vptQngX-0c3pJPhrPc8I7x3bT59_VpeoQG8rd4b7smlwD_0rM8CCJePrBDeogT7VQbD4CexPUmFLtHtMPgpfaIRRZufhWkcSoMjtiDwJYiI=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_taYkbs-kTmeAi95iJzFxwslnZIZoLwcjQ5_b4NsweR8k050IspAwdBblCFPVwQJ379Y0qLcWV4KfXmBk0p1LLkdL7huTJp0eKnKdVH36Y7v2v6zXNTTsGdcnCaiFSnQ9wN_RUDOCQIGiUqCQcaR4CUfp0yYj417ygg9HzfCVwx0ruTKa1FY2kjpaCtknkjorDPF7Q2jTOTmheArofOqKcAe0SUknmpvgpRBgDaC_W0VZKPEKg=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_uaJrqNbHbf-OTfpUCQxBvDLXHeI8ikPrVqXD4t2ly5ZlEkNeCJnVUwUMVFlcrk9h_I_br3EDz29cdyvjaFM18y0_3zFZvuOeANDuYdUXoNfBlfpRO3snFe=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_tPFa5fTS87KK_J52N_b5uhj2kQ9J9Lg_wjWnKpAzmUWUOmL70_p_ZqCs21PNS_k84oy-1XufyaX40Gw4ErSRUhqWE8_zHoQ_UvLyzQaaSUhHatptM4VslGcDSNG8ElOQLhxEIjbKQavitPMmClxtLoAlgyEMZsjnx-XQBgNaIcE9J8JuZojEo9PU_iNCmC7xl2FhQrgP7UOtJQnRNbDDVZpc9Z9bvaIxDlqucJjOc4ESXleAI_OV0PKsIsNtcYIJaPJLIpfu__YhzB1Y-r233A7vbjWhqrkChMpA7N=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_t_ZO4kFuTsDz1I7YZIYFaMuHqmR6gYdRZjKH0vRUVBo9tJVRgDCVPAK4KUCryAjjLKgS7xyjGmlHtSP5eX2NgNZ_897HovBz2Mx-RTbW5oUChXRNW1v665xdp-egMHHNagwxUl9vdV7uPljs4aJcJYwmNf31mi1b61rH5LnysFyrl6irjXHP2wNKLzc_ORNkNU_NHl9tdcZAqcvCZRF2I8bhc70Kw_RWWIMQ=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_tuGN_qOybz_LnO4LY00TPVz5ONazDDhIAEA6AqnFUoXVS6zxfPqu-KyVTio0g1P4S83LDabIHMGg-Hcz1MEs00DlW0Q5IDg7kcCiVCTC_VP68f4DA7vPowxSIbnxbNH7iqjRF7ncTuRXzZJUGVJM0r_iRZvj7-d5oROjMKJOSwVkxf0yzVKkyQOEhjH7YrecIW13AJ8Vb-1FswPwj5xdscpnlkTyhcQKXKhbZHeSJm7XLDG4XKXMduK2x3csbL04dGFItUs4KyqdSklkwbEXgEL5-_Ri2zJi2klFedRwSzGWbSASBmW_GrY6htCnXtqBc_=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_vP0iNSHZ1P2N526ttKVf9Ri8w_Cajm5i5mB1k6Jdbdu9fPC3R6HzICU565-GPYe9fUt6ZcZc3MjuDH4A8Fqym5cLGApDf-sDsBqpig5tOtFV7vHhdLdKPhEl8XMgsjzEiU=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_uLg3mJZnadsT5MguO9dec3333Q_8Rx0QqtzlTRvtf3t3zI0aroPc26H_DElys3P3pGSTVVPHH4zKen_te4f7OQ23QCugZgKnpsDYqFQwA_2_xzsZWLOcBdOiBO5MZrR1yZN0M0SVIqcY4b3anY7az8neygyAVMjkl4XLUhtFbGlUzUzA=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?!