![[;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_suANg3plWqQnV-3aTytxu3xjprkuM9Rq50N1Vg06o0hW2HdS5RJWTvV09xNH_UXdsmJgXbpDznRifTHYNEJokCtLtphr4bVZzwjz-QLJ_2bRjibntyKvAS7tG0TitsMPyG3g3buYn5xWvb_dhXZiWZrQqBpw7TrDBYvZnYt5MbAvMHFzl90kAbKEW79xoVShCSzAkUuUS-1e6n_9vuHkzFwO4YpFMcUdruTpkTf9PbJIxNLLLIPppvzv5V0doDrAs7eNQzBxs84Ml0jVcwwcNDF4niE8QnpmRqHLznxviH4DypHWyrdVavX2hUjypfjimA0Q=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_vpY0y1nxfVrT6Dio86bW9iVIAl7lvx0LGrCLl_GuZLPg9mUB44pRh9xM-KKy9gvm6PI_pSRVclEuRSff4HyCpPkXTXpa_tNhbL_HK2jI814K406G_5imAQ5VDrQXt7IbU=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_vbSpknBmwCiZdLJv3h4CGn0gtQnpRVkQwFwWccXSN34rrUCaTIMFm5RxJt2KFOxBmd6yzdRIDQs0ajPRP6-9NamkKSK0kAiesraP6eBcvvShXXXy62N9uHHMiHckoZw22tRKDrkPi792OvXxf77fxdtKIfaXIaFVfF-0Ok5mseysaUkW4y61IKd5g3zTMpN0KSymxDZHo=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_spB9VwqGvTh6wwAPVSzwPFzDWhqLc_CKMWiMOQUft7k5qRmCwotkPPV1wYpyQI9uzLPG18OScngc6UcT22NaXRjwR6GGDsLYF5wUehNWs9Sx04h8c4rd4n7Tgb_aB29c_blmcVIkEyDbf9q0uP8liG6gGQtbFIl-IEmEWjPwqjwQ=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_ty5_2LdoIE9xIIsyWH62EWJ0Jqo-ZM-NbW5YjP2ZjhljuudS8SRzDVI_HDGtc_ixToCT_3kLAxay7l9ZcquGS7gsMrLmEn7s-RlVLajA=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjc2xPIDAfzvdLonKaDthn03nPbJMbfpjcdxViPsURrya-w8gRf5wOlwX1JFDJ8P093xT9JE7LZEvRTUauU_XH5n_AnWXrjkUxdIbsgH1_nlEltBQdDjap-FIilSSQLWTvzcg=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_vaHa_PRaL-urvDR3SDTBVyX5rcuIQsN7UELZAC0IU5ln1FaOVHDv3NufAf9VmktL_63F2qUvYf-ie6N_qk=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uT7sCJx6MfwlbcFRLY9hGHx2qOsSKBC0mGkaxlO1RS_Qlmdp6m3JLA0inp263PyYoVTXLISRf7XEzkALIJJQw8uSr4ySJ9baPJr6Z5VA=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_usCVuCSKg1hNucKFkmS5Gdr87wwlkr1d91LbWdsv3_wT11dYQbtvnVnkj9q7VHQyyU7eSysE5P7m2D66R3E-I8_sgk0c_tjBOTALsueDkoF1ILhlpyIp1nY9ngYsR2S7-yYEVnH2S0otHwPLyjIz0EPOH-gGu70az-ONJC46uLZmbEpyQookwFeJcbHg6g5lFp9teaelG9KDRDkNYSU4-LOBDrSJT9GkZjRCan2N-5HOFYimic=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_sWsLFeTRuJo9Nox27YfL8OSnIldreQfl5uT1UvCaCVuNibBfhW3kjSNv7gSULnSHBz6R5i8uikfcv3OizTgS3ftHFlDX2oft8XLuSx2t0kKRWF0EIuimn_OuSEt1zph8dUgKrfg2J0moylnPpf6N7Y-0wmJQFd16rMrsqXbWJ4RyoP05uN3E1WB7_Qm9MyzmYGmog6OMH4kpfLDq5HfmCnfmHpGDmysnkVG0feFL0UdGwNZ-k=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_uogXCZjQ4rJPgi6XtF_32wnluam8so5_jBi9ZPyXMalqypyzFKS8OGWZZIP_SinOfLJgbN21S7abnq-OzuIxmglOfTGYwtlf7rek54dv0ciN0vhXf_rCas=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_sDrZl-XHawnB89atQCyIwvL8Y-tZa3qayXwl7WA4Gi3OXHwv9OzwKyk5D0IMAIVzG0D64gbFZzkDIR-jj7WNRRd6IDOdQaU0kTX7IebGcRNlilC8_7KgBEBE-qMZbk2raQPvOr_TS8rZkT5jT4F4EXb-kIpmvvg0LaanMeXqWR99tFY3C7BFBd4nBqoQcxNTIHx6PFD6w_Y3V2Z309ryioVBa8NnuhyYkMVsIEIzIFYoxaEm8RhxRI7S_7SzpLmJtZxX_ygGKDRw1PaV2B6adlWX4r9dozmhxLlJrR=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_vl0IdR7_ZbChUUpf3UdM_vlCH0E9xz8pvjZUs-6yiuZsjksfb72nhMBBMN5gXqKzFebO7cN75HDd3mpRIM03xeQ0TzOyGFcXv__pBMMsa7dHm0sb-CD1N7bIXCrG2bu_RL5lePGph6aq4rnLFmFOO0S7mlVdQSrgXS5RW30b0VyQNJds7i0NV9UMUudYNTHYQvipUS_UvUFu5Xczn1qjG2bYRB0PHsho7gvQ=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_smsQ_kqPfYG0Txp5sj26Mmo_yyegcR3rf8SJwJHGHX8jiFgewgALpyb42XDSmoxpSf1zDamsmU6iYlfV8EKX839mGDcEA15HfzYOIZUhyNt4XB23HgMm1Rflsfjul7wMMhA-Cy6NS-BSrBgxHXa15dHVHxATV-NRAAwwtYZ43GLvqegmm2X2TD_0JLMo4FOeEVvmsFLFG8urk7zvimjL_ueTU34s1ix5BvCUHV74IqpP7B0i19-8GxGhLHz1ompSDlWJt9l_651h4vNNS9XnguxGsaFWzB-OlpkiwOsSiq5NlnwRv6oqlZfiei-VTAo0CS=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_tAIfl2eRiXDlLVIcDVqRd_8_bZzTSEDwwHuXo1Wnqlyz3zLua3tSyQMB3EDRAsaGwh2ZWRkxvwHg3sbmYHpfA0B4nbwR4kJsxEvZIZIP7ceFh_uyknhwHRGvsXJ5G0UcPZ=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_vVOt3hq3P4cOjNv-qTueBTEu7GY3h94R1TQUzav5cm7dY-M-hb174PEEJDDcq7dA4Qh_8ri9utgNwWRhQOHw7Nf_IC_a8Fryj9XomoXD69srHygWws4F7m66gseKCysoI_D4rAzZvnJrpWbGdoRSQxwgCvG_dt6mWplQZz5JKfKHNibw=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?!