![[;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_vcvqJilXoMZW1_1Z5SybUVCErVpW745emtpSZO10vm6jOSbSC1ozBaBUSSrVjlHorV0K6ELazWGH2E1dAlj6C5H3fl_TwM7P_fiA1xzpfcVopuMQOEolXUpaCQR8dZ6ajdYhhi1pQe8277VK8h1u-KvwxDZNxJnRrUKDlMTI2wiD89Yi-hHsrQbDXhPiTi8G6uWMvWSGeH8UDbAfOccQE8BnF-5EK5UAZBxv7SZC7sVJNbYnD0gHe3PETT0rlp-QtgnjU-zlfAZQAyMmIab8BYeKyzNXpu2jiYBZfjcXbtcWuS9ic-Hu6djSDSavTe2Gq1tQ=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_tK5x33NqSM_nFb6P3WV02V83F5uPGuXtdJNlyg8dnwvSvrVz73HR2e_FJJ-zZOqFNczYyC9YBu_Z0hZXgLQuYzf-Y8s8_Ncq6uEgkXIdFP_GwJD8urd-pLDV-svO4RMgE=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_sgUu8i4gpzjJW7MzvBw6DeczsU0VpcmDnBafW6Km55jUstyACeMPZsO1FJTasRG14LqgVXMXjjrcXoydYj2Q61RjHWZIkDsUnhWMpFLmHwidas3Id_5ujbUfP5tyvBy8gzycidWqcf5WJnhbLf18_lTrHgZmCA2pH9Bu7ha5iuV4lMXnoOpJebeqootgfExYAsoOzL2jY=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_vm0UZQuGHsnd0LTUputcxWOlkBW15E0RdfwyvnjZZjhty3k_X93yPcJrFohyCbrx4CyHEA3qD6keGP0KxXdzIa3wB2nS7NuFbD_V5nlnDQFxBczR1bLVGpitf9ytcEVieuNIWGq93S5mffVXfKJgKDQ_cq921-HWV-VNFT710gsA=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_v9SspUCGNm-fkH6JqUbT5GCy8howtvQ_AaNf1jyAMoTtEopZ5Zdbb8K39pnhK2_3jYRU08IE7xLwYV-DeaecpKqwHS6QQg7MHIKVRNAw=s0-d "(1+x)^{m+h}")
é ![[;T_{p+1}=C^{p}_{m+h}x^{p};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_udQ_R9wjsSKiFzhCuT6MaL19wGsmwMQd3jS240CfnoYYjz-qoT1c9MpsLmRZmZHTSnXcDvvIIcfD9XZ-ASoxXBLTrPuttZGkmQzjH_gF-p0vetE9LF67qkbCGSHXPKoqFbRY0=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_vyEvRzS35TF8Jr5-ec5F_J7a-rLa2WEZz1h3QGtcY3_1lhR5C8Rs6AwibwTHqkpKSNhlUl73JCeF6AC3Fd=s0-d "x^p")
ser ![[;C^{p}_{m+h};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uIVa0G-B52pkdIUTYHl6eoDnFAKdMgZ-Oy_9LafZDS3LrCGJkxBL2CxFGUjUm20Xc4Qcn52Kgw344Pd_nxap0YuXtmBfASpTuwYgz2-g=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_vSQlwmE-oNgsYgT3gZVRQUwKhsPjK5k-crABUr3V6CFylDz3EZwPJelPVDCci0dadFTcQ01JiyGCTp6Pm1ZiiglP90xPQ8W3zjWnZ70VpGFzSrvfSdm8vl7fjwPb2R3TmOPrgYUsysWZQNjZ4zz31sPrbTGrStR30ou7g166nRpofuNM9o1hK_0WGMR8ghGuFZOPtswTJOJWINWz5auvK5K-XvKDKl2F6rcfcDVjQIk5g3Nwaf=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_sr93MZqv5gpNEcvcFJOhVbyvxqieomHIGh_UUzGhOFGTYKZo-9ctspEhJR-pZXINFRY5zuB6k9ZXsfUKdVfKV7tDUTW881HgUdqaU6lt_1dt-8W2NBEhNStfuneqc3HXLRsCa6ANk3_EOxETmeyQJBUOkQcP5gCHumaCt7wU7TKiDPGPsQbhc-qzNy9WlzRAGuErhnAJ4qIlwyXflPeBJ40enDJr0SXpKSiacYlU3fIZ67nDU=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_tIAfeiggC5PtwO0JXirsgH3QRMjiNPB2V5z5IEObZ9ZWn4dTYiqaBeE7-MemQ9M2TXzDSYqUUGWKnNczi_APePJk6JUHPE3j0HpZqPjntSyA66wY3WYdjZ=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_skF5xQZt2zvhtSepGvfDnkoxbznQNnRgvRv2zWXZdSXDJQNbAgttC1KWu8eVDOel_eLTSMU4MdHYunhFLvo2o9_-ybRqaNYpv4DQwsi0jqeNulTBQy_qiGbDom5wIfD0ud9AtJ_C5HoXFWk8Rqx0s6DF1kXrX9Tl735YSot6g61Klda9occ9EAfVCtzVdWLLNhq_KdgLUG7dqJKGp7o75GYZTmGeaVWdHKNwRV0Q58yHP1sexBaX8Uex6RFFXFNJTaDciMVgnokz4a4Li9939zP0i5TfZ2LDRLsyFt=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_vGDFUgUE-pZ1pvRxpqPSlAk81Q3PAIv2MXpVcLVMefku3pnHYa0n623VwYtipz3cRUGjKeuDNsfDOwO262cwiDlWLMonxSs7U6DHJm6SKQzo3xft0RMUUXGVX8DVMoOVDoNSMjVJJtElbnitVXZ98RpQLeloZmUk26gBIfjgJxqAqECI2lkHgcI_RcS7g_gByIUlzIVQ-ODSnf4DwpH1C-wIYJPy7U7NtTmA=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_uvSHvOazx1CQBrCfF-mf4_EJbzqFgcpHHjvJM9bfz6ziaB-XZhuw6HKz4boshZv9Dck3lgNRPPapRkOu0zxWr9fusjfR1vn4f4xqli1Ojqwl8cQfexKnpBR3sBFSmb4qPY5vijDkZk8Dvr7DZWyPRzEfUE_kV4Q5n9rrKkDDxNozqfyl3pR8RosjbEYVvSlTiMrC6NVuu3--AmE1i2t2awltY0mhNywDbdk0EJ0k5RaK7KqHZ4Fy2GgmhWv0rgnOX5Cpx1eAY-WlyWJlqzCQ3woZzXgV3RPRqxaz3kKghDYH8xxtojNwKKpcYAKc6wel2I=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_u2I2WFEV8Y1YoOgcIP1_PHl6YeRmNqnaQVV9CaaYNcsjnSgyHGC11WImktFv9eJUci7BJ3sHZzgapw3f-vrvv-w8OvlhZ1otcMXg2KmLZaRXNFtA0HWCyjg-5RLOrd8fMJ=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_vkelnUHJB870sJWt4nVcv5anG6WP4s_OXrvR24mlgpSswMY6POFljbdZohTP3rW-Qk2WDq74aczwLsQEMgJq3w8TFTa72XfIGaT4p1jFRsAlBtW-6eA1XbjFTE5_GzWPGwO4FdUtr-TNypKfaVYQUzt8Dmtbw6YS_Vep5mwY4BKSX8Iw=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?!