![[;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_vSnXs2-_JK-ewThZVlwyfzF4-rg49P8Danr2zFeCj4gZYPB88T_aXaxov4ceuMPIGsGXCVXiav-4fgSOtjsEDWx4QMZz16DQrt5x396iyDGbS4GohfFwBexlOrxt_0pboFF-bgY_vFN2gbaDisc7pz5mIN5RxeE8Oqs3EoPmfbVbjTPdjGAmHvpRRF8gozgDam91Z09E4euCS43DOu9UYvJSBLxNGnTO6KYqAjq3ZnX-6kVjnkBwAw5A8Iir45xrLJfLZPjiuUl3jid-icGZYY752hVLAvvarEAK2At4yMryNU-mZIL33C64T4aQvQlE_syg=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_tZIlwD2knG0bz4PEwCgfP0WSJDAILByStO76x9q-0ZQpPXTQCCs4luIkyLbzRSYmAyNOL88qeVL9qSa_NMvk2VwPJFxkxtBS1KkCRLoNIj381wBI6inTJryFtZusibLYA=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_sLl6g0IcdxAbNfALxnabZ-2HHxBIBbgGAOBWVxkXIVtnzubXOvoUXbMBQuMbgkNMmJRLa_o45OVm8NjAL9b8_YLqkJbQ8tiEgiIv-EPlG-uCz0jg0DzOEJkahUY4VtXgF9aMiHUkE9MY_QyHSy3MPwZzJmavxf1buq3HWztZVcieg0gDyfZ0l0oANkd--01MB7ODz5JfM=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_ufy4mafjxu8BE3OsqB60YxP2uNIaGBhLyNx2Df6snWYapYm52zZfi-NSLWbmqszXOdKO89xAShDqA659PIPfYtbeSCNkc_yct1jHVB674SEttBCs78hcwtk1wqyKL3zC4IREBLKJctC28GQY-Dokkf5GvFgNEtA3N_KAW_Tkl2cA=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_s193RH4A69krwIptKAQ76grX1tlt5wsGsZ8e4M4k6L0qIZGirQL92JkZKgvXdLW6dj-WVSFLMKNgh0QgE7SAla3emvmH_zfUWvZxAmDQ=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ukOht3gKYA3kUCbn7Jmf4ZH_nUD1b7N_1mNd4ocn5aSNyXSP4Yx3mtOJ0353EuprT5KmpFyu1noGC2vZtBZyUnjip-54VGI5-sMXyy-eJo3vrHmpr-cskNBaUfa85pI5zzlkc=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_tdBVdPyuA6vyNcML9zFdShV0bkfyOWXQZk7dzba8kPiaH2jth-M1003sjmwsyGFfi5jS59hsQtFLhZv9nF=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t20kRJCUM-RqKUk3jjlpnC48VvLO2qI2l35uJGHQeMi0tASxQAVoj2gepgW5ugBJL2UcGoeDeDARP1wvkyCVukaW5K8-OODrFobjdteA=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_vxekq3dIGmxezw5RvJD1XC6xy7YAKm6koiN3mSkMKdhYcPamx4jgEn3-5N5hNEHWBoLPJma4SomIVOE-Q5gCLTCixQuXAB-rP_af-JeF7WqAXkE0VoTEb02ouWpkx0fHjYil5nLnxF_c1G3tHwpWfgCsKYwTJLuq_qvF6tpg3v4EeOG98sOc0OYeGL8MYVag37yT2OpfrlS7Fzw9Xee5YN3IqAMPpboJEtM9M0XsJhSOAXbYHV=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_vWbwrX-Ya6c0V4fVvYKqdQmmkMR2PjdsRPyGItIJWINMoHPEPa8IV1wC8rP8J-Ma00Z3MV6EUnI1nB9Tfi6Wv1x_1E74EZc4yTkLqPzM_NYA4knI7GSWaf6kBIvdu1f6CwNci1BZPQ6waCmq3Y8A29PSwbkvBGxFDsuQjXkxLfXfrnXsFwM9Rl8E5YcxVDlBKWOrYV-4u3p_zng_xG7iyqg4BTIZ5QsLIqhcEAscdEqCt4BC4=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_tYQkK0negM1xvx72NggYAxGYovPOA4d3TsRqonW3xXS1m3tsVuOTjD8UoIWeSDhlyJ4u6Gfp7YVQg7lF_l-XlpkbNFoXeXV6T6KNA1MQWBKonuLrNMYQ-O=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_tM2n1fV2Ru6wprHyPqJwIlG0kZLOBosLXnuszPXtVgVqdKvg_amzVNQ3vPFhokXKkQCMwO-1y-6I4oATUujnMNEA5Y3Tf15Uf9dkmZbMjGva91wCitv32cjx13KBO4fy44fYYVVYqYb_edTNxa1c_MR77L6WgAkeCvlqAXREhSH499MnvYiYwPxjF2nli213PLbrLIDlJ2JkiMpzKNpCv_HRI2qahAWCVRATXm77BaUDdIjh8OBwoQ7llhTXHApTLBVxDL9fW1d_lY6ql0JJx-_JwpMfyfDkQtOkno=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_sjO3mjWlsRBwES-ZEj6E9tqnow4yBXWJr-02vfb1Xhw-wP4Bu8mFZB8pBS2juSk7sSy60NQStTQf4jMz5yCIKC8eqaRsdz5W0T43vli0uzZVIT7wkUjqAMPgElC4M96nfORT8p6wbctShIfPBMuRA8ZVO6mk6nMJ2R4YZtJgtayRF4peRRWoCIxcpVfRhxoZr75QNlyKXwhCTIpQVsWzKdtieleL2CMQXWig=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_s6PYydfHKrueLQS2Uy4ZfOR8I4X8RqU8G2dg872SBxMC961zOj0hlsQfP6fRtkCRsBLPJ6FpTf0kClZRwJglw4OY1pKSxGmyKxKQqKSpwFQBryiwMcvKJ19wnPrQk3Bi3IijFiNVvERC6mvtJVxQNS7FiTGjGFeLwurIITosjY5qJDcrogspqXsJTS8Jr07BfNYFxp-30LE6il19mUDleQnHk9j6C5P1R5101hDXmMcpRdznEGXLDr2R0Tu0xqk5lYTPnVkXe7628YoK1BqGHm_wN4JMQyhtQ5Z-z9CM-jukV_NkdJynuIQ-aqyeSQ_nT4=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_vbP4LPjDQy_l8sjCYYK1VJ_ZHUXju5Leed0p8JyAI_M-xod9AZBPdjLWldb4K4YwBhNLjtFKKxe0qJAmwyBWKic4QxMXGBSdOhYppKMiaO_K8F7KO1lU6z3Dmhe7eswKbq=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_v_8U5wDXp6yMPktgfbRmxODMV62Zucu7FBLxKkf2fSAe6DQG4YtlNtIsTck8-zCgg73JXqdqs5KGkK-F6Plf6tBiKjjdpDfXo75p3OuT3yKIybGLrnB0Dr8GDodWbbY2rMmKSe3JJuRYIyedj9wkVnMmyevv08V7WjLVOnP-AazNRZMA=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?!