![[;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_sRxxvjmgt0LMiusv0MWJK-gO_uj3-XLZKCQ8bRW0TePBlcqWHaNh4lXNr1I7SWHCvAM7CBVPh_SEfU-aKALZLAdsH3b4OtyzvSnAnc5rZP8NHBI-3wIyLysW29d9yP16P5cbwPk867DRXUFTBRUszEcbdbpjUz5huIdz_8ATNIcTQrb_ZGN1ZeeczwfKEW69txoF4Di-WGmQBQjtquZ-M6swFXMNpGj9f5QdxE8hoNlOoLiBuFG8_uy_pw1USdquUQLg04NHsB-lC5zklHkFyAj6N-gwH_d_tBsaSgphrOJ4TeJyuWN2sByzrXpANgUDVrrg=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_tptCVZvUiC0qibekEgDVYtMRnNB5zmfEm9BWIvWMc7WAhqME2LKejrOO3xF3ND-0b5YDh6lgq5yVv-jt47ZegHJyRQjU-BH-NO1AL3QZj3NodiWu4xs55HHP511k96o8A=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_tza1G0zab33aYUr9P7qe4eiUThh9qQQkkqpeYry3uZcZYONXwaTW4qOLp-yAxwj_FldUc_03MWD6iTWIaGZgLjstB0R2z_Coe2AqWRC-ZoCSPHXFbsglBc-QlnQd9TdDpgj4oleS-lr5_DHUWjizY-Dqa_URP5iyDDImkl1o2RuZ4KwONrymLMwrMdCx-cmR8mR4pIGkM=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_uk9YqQi8-MaoRm5Bqx-yeSFopmCxhPA-GJsK5q7yAf82li5m0QrVa8gU0ok7XlaDZhcbkexhM43lZsU1sQA_2LcsYjU3Vegg3Lai0yIxTMbiDKU1Lw__ws91cuVTQlMZXcxj8WoWNUOXaPEPwh_kAoV04EM3yZZTLqb28MXyXy7g=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_vqFQUuDuvLxOSw2JvR_-G8af6hut160-kSWvKwKzI1xXx0DiSupHQLRgQKhSX6MHHolv1nnvA2qpG046t7YPAWvVFXlxXCNXnianl5ew=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpK_j_2UKY3ywNxFPu0Zlq4dvwE6uRwgcNFxUQTwYOJKO-yfjxiTZY3bz9q4egs2E_OTpgIVg0uFIXMyFbW2EtRsqbLua6K8OxonxzDIn0DdSH4d2ZYWC6YbZ4tBUx70O53RQ=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_sNlYHhLNQzjnZVwckyoNFeYeAXtMuR61E7yBf-uGOHshlYXODTsANrFxoqdSBOJYZs3E1-W-yZ5yE_gy8w=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_twDHFRWDo-I-lVqWmldi1L2cg-k6ygvtXdFiRedHXrvFIShQYZiuaKN9ikKeNwKrzaxoTwFzjtQlm5gwN5egOvuFegZj86UHI6K_UL6Q=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_uuM0IMY8SXYPhhNlk77ujn598wNRGVAnhLnFdtVGz0OsMNCBLq_HglhD4jhZ_aXg2eUn5Wf5v6Yx5PyLOzFIYKWi8J6dm9UV3_n_lSbyw7p5h_YMrVzNRHJETOAY1kaUXlTX97P5BbrGEtNKaZIk4yJj91XadZRDHapY3UCZs18qHiG2gWA0m6YfNNmzHby6YXBh-dArKen_-4puK9NqHNrCvLMY52D2khmGBobA5QfJQVxvy6=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_tqepPuM4KginCOXy3_19XBsSHfUIFr_TOoYjDiJllos8UiLKMGQpxkaNTxY9r7wFHf8O0zfinXMh0fIfFFy0cSqLBLsMZZVzB2JF1UJGFh-2Z35Qv-Wv1bAVquyb4kCxDpfqyogyVyY0WiHh1vkPE_Xf0-aIZZa8CFY1nFg8e5c5nDJD23x_vmiTKQ9lfMxekk9c6suCCH_f5TK-DNZKuZ7T8N_i36Vn4r5iFNE55N_N2ksu8=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_sYdasdeZITLzI75qMYkr62FZswWkF4BchRQsZGpuZmBi5l8iRUwDalWCT-NQ5LNzqBf1HeFOQkD7HGct4hUBrxkc6m266rWgM_nV8trg0Ns8zMxsSqJis_=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_uB4tItba9NahPHgYDHHyPsEHH66axrYF3F9NGz0kaGioNLd8PG3DTvI8NMNVBULuHynO-Tfh30LPoWEr-nwyJGxMMBD0uMdClFsf2pcv12n8iJmF5_0_v6yTz1ouJzQaHJX4X1lSBV_DXD3RBmMbGJsUcworpBio1OuyzD0nCO3USY9mrmAvAsHujdvsQ0ju-celQLMqw_611aAgm5p7V-Fsm0R4PHdL8lwGHW6MI1XOICijSz5tt6w06gLDjbuEtuZ4CIACAZChHyLqsdI2FbBn8j13gTSB9iu_mP=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_uZcfV0zTBzhhjKFi7iejTakbDV2_4wk4QKwYzC3DFfLh1p-yLXut557fCEy3D9M4L-MbjpGJUcOzcI4_9iM3uNlSztZSQGX7fkNJnZhDOKe22epnYBb4kGoMtEgm_0Ynk-acueJ-ZU_NT3ziZ9VnplHGJsU9beUKPDNLu7E5ccFU8dPxEeYp0LS-0alItiGHljot0vkzUCSXUq1Kx0zJyaiXMLLvfQGecHiw=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_uZXRa0NqGuoymEwGoEiEgtikOZw6nphfayptEvjkTL4M0zCCE0tG682cPlipIzJ2I8uQ0bCv9pd3mkK_CT9asLNk9uBjmNdBbrukZ1h9jnApxAMjK4CM1GQRLpGSFDcaWaH-xS7oRTnAQFeBkwOY4nM3Q-JRqptYTFl73EGeNS9o-UYtuxE69FBdBydhw1q23zdvgn8rq02fVudoqIASYeyBmG2ag1kI96YxxkwphPRIs3POSCvHQ_HUyfJNquoXfgPeLrvc_i8M8oUl3BNLi_U1_DaEGPgCn610Aed5PZGlY5UPNkzTrY_fH2tyuxRqde=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_tnmpERG_20vZCT3X2myH6QdEZYGyRHjiXlrNI0pRTm3tIRDto9tBFsC_pCdF8BvVqV5Jp5fDwIsh9QxCCiQ3GTp7XvLUxPIvsWa9wBnqRChlPxYwHauhCvstliiPiYM6dE=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_uRJcI_HE9s3RnXDhWUrH_nXobYO5i0aCKpXfAJ6IzYs2VktZiuE_UhGJg6JFqxbiSMaoavwqqz51hmUUPJYRvmUOXza9TW7y_X3iOo7msVUf5JXTKFqDjOxBHYZpGEvFmiGEhJ6Osm6jgefSb6rg4eOEoXqb3iB8kZ5O7TK55r8fAxHg=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?!