![[;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_syqJe6-BS2ld44xLlmTZdRgX5t6RIOLnXOgb63W-c16g1Uwi1vpNUeNfvElm1mVPSea30kDHC3uWP0HgHkfhBGgYoZOmZfX0idWNtnrsv7NDBdta8QjBymxFlVm4IlCR0FcmRGj5al7Gn4eT8jWKks3AyvmHEKFJYHm25cdTIUoDRVFjNmu1OlArbLGi5kpNSXkll0jcm0qc0jupKl9YaTc0wdUkxseNyydrfSk6DupN9wU3Ydv_umX6ANjnawVfr7HDEV_L7eG6qTW_U-OM7f2y7XEm-UMai6sqR01WuxA_WrEPJU07VcPWzmR1U5gFD92Q=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_vHmvR5aqiPkWkhRDEGOYANwG4gVRmC4dOH-TFCTR82yyunfjWJVFkZKtkXV7xHpfcvrkNmIBNid8a6zbx4aZTjAJ5_nZaBK2LPpCBu0qOSAF2wt8kaq5xDgriaBafbemU=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_uXzRZB_Q_drAhysfwr7gvjdrr3XzWtn7cmKVrD5IghO4Tdiwp5n8-6KRgDvkcnVTKeXKSF94mosAhrspShv3Xfp0u7cjLvSG4CMVwTNTomql5kpYhwQTjApwbjiZwS-wyKCZnb1c1Z-q8k9pdESC6oZZ-h7Qt3Iz1HV7w1zuvpVKvWa143nKg9twy2302Jm7VhChMtjNE=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_tPe_QgtPe6U9s-g2SO68jgbJNIi1dbauFSvkv8mjNl-Ucb2krxco6bJ4IA-EwDPfRI0Qu7IoORqDPlnpWKOwcaUoMPB1vx_riFaXznxxofBabOOUg01vuRMbzzRrLN_vi2sX53SNttx0YnQzbe4dVC-XVqGU5-XbI6Yb3fQ1ljvg=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_uvxey6MriEZ2LDIMQ6RyMHmDkmI2FLpSgjEnSLBAt8nXx-knFPGFhZNLHy8Y8PL7wBd9avyG6bf-ryrryq1dEkrdxi8MGw--NxvYm-0A=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_uHHPblESe7WV5O6DcVJAdl7tugkaooXDC1r0q_geziTUKXBb7a0GEQmx2gjupP9YUBKrQltRF4YGMVxACqtio1DpSTCBayyJQqfqIcGd_g3tgUiV-utso_e9ooXApda7FTkr_ZPqvtdBZSiYUCRZ65u3TxGwysLUM1Z8JAjq4N1CcOw4yKMblrQPr1Jay8nSy7g1Cg8XPlvy0tOyXbYGkHDbGQ8MlLfKy6ysfCyqvavk1ClJhP=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_uLxDDntVsyPmObI2FVohOW0BZMfMqNrBHxzVKfVD9rD_t8AoSRp7YtLDbrbmIu4tL3bf64zaeSRDtnUjD9jKDcSNV9OQW60vSCfZMnUaXhbLZPQVDkLstpT4umDT8_cVzbnQuxWCHwncSgLpc5MZb95TvtVPuDLZ2E_cESXEJgymngvjRYJabxN33qGj4DmKEfd1iEiXlnxTYXN_cch1eEjQ_ycb1dqPY0g0c9aVXS7lpplSk=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_twcCpg26w9JPvnYa72tMXAQrsRWM1lsJ_6H43IPCmRu_BMZwgG3PyNPcNI3nxpUy5QzrODw8mM14vOKjOo_9cIDkto_hP0jX2_0khLQ0mi4nZ6prYk3vZ0=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_veGZq72FPpYP_bCd6m6ZRXjhhjaGDVfXhF5LE-VHYz1F7HkDYJvkhUtwtT-2gpdmwVtKEPqoHvjBcdNsgbEEQ5Sxlp-yVhxajXNc8ghO_wKtdwuXua3jBXTFWY4l5i4xw6Mj1h06f5-MtJcxDd4GVsiUDaCVT11kSDr7XTJSjPpm8ALIGNmNrO6RWX0GoMbK9is3xfhXfkRpYbLKBweei60rllQDiHgXp0c2C90BKeitrc07qHHJ7gIzxjLupbXFFL2817rxMQOkvcjMJtFcDpMeVOS6JIBib7HkCF=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_tawNKf6Xbdr77OBOBb14FdGFVojTzjR2-RGYhHMk1x5omwLnIWKZVYZsCPLrEYQtyLEVppmpMDyX8XSG84bRr9qVz-YdFuIahzg0F12aGNH66-DRmFm9B4m4Sjapq5N5t0UgnZLfnSj71HPxJ8aBgDNqQwhb3Sf_ViSXDYuRdb9nHUtUtoiFGr5O3h5WbeHVNfV3PwP_BU0BctgX9Hxe1WApTJFi3aZx0Bqg=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_t1DJJRnjEblbtzMXOfdpitUfXLY6BaglgzOp6goUHqdoy1CRdozdgvjNj89-lCFAY3weCavV30sLIBK-n5r55Z4EC27cLHAsnizJbmih-xfYZSMOLbXmFcdE3dl2NIPxPf=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_u7ldyTjOfBwxs7i9YrPSh4B04S-6kZHhzlIvtZUYqVpHNnqAmOAh6QKM3QSTAKstgJ4WaCak9VSrdLxkFOsm8tHH-72aGNdQtjp5F-yDT_njkRPzRdm1quhRd6A63IfsHiasmLEIP4Kh1rG6NsVzfCb1B1o4olDu7KrTlDqqfNbSZUNA=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?!