{"id":2595,"date":"2022-04-30T15:18:46","date_gmt":"2022-04-30T13:18:46","guid":{"rendered":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/kombinace-s-opakovanim-zakladni-vzorec-vzorovy-priklad\/"},"modified":"2022-05-01T21:34:09","modified_gmt":"2022-05-01T19:34:09","slug":"kombinace-s-opakovanim-zakladni-vzorec-vzorovy-priklad","status":"publish","type":"lesson","link":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/kombinace-s-opakovanim-zakladni-vzorec-vzorovy-priklad\/","title":{"rendered":"Kombinace s opakov\u00e1n\u00edm- z\u00e1kladn\u00ed vzorec + vzorov\u00fd p\u0159\u00edklad"},"content":{"rendered":"<p>KOMBINACE S OPAKOV\u00c1N\u00cdM vyu\u017eijeme v p\u0159\u00edpad\u011b, \u017ee si vyb\u00edr\u00e1me prvky a nevad\u00ed n\u00e1m, \u017ee se n\u011bkter\u00e9 z nich budou opakovat.<\/p>\n<ul>\n<li>Z\u00e1kladn\u00ed vzorec pro v\u00fdpo\u010det Kombinac\u00ed s opakov\u00e1n\u00edm vypad\u00e1 takto:<\/li>\n<\/ul>\n<p>$$K\u00b4(k,n)=\\binom{n+k-1}{k}=\\binom{n+k-1}{n-1}=\\frac{(n+k-1)!}{k!(n-1)!}$$<\/p>\n<ul>\n<li>P\u0159\u00edklady, kdy m\u016f\u017eeme vzorec pro Kombinace s opakov\u00e1n\u00edm vyu\u017e\u00edt, jsou nap\u0159\u00edklad:<\/li>\n<\/ul>\n<ol>\n<li>Vyb\u00edr\u00e1me si z velk\u00e9ho po\u010dtu barevn\u00fdch kuli\u010dek t\u0159eba 3, 4, 5 \u010di v\u00edce a v\u00edme, \u017ee v s\u00e1\u010dku, ze kter\u00e9ho si kuli\u010dky vyb\u00edr\u00e1me se nach\u00e1z\u00ed pouze kuli\u010dky b\u00edl\u00e9, \u010derven\u00e9 a modr\u00e9. Je n\u00e1m jedno v jak\u00e9m po\u0159ad\u00ed si kuli\u010dky vyt\u00e1hneme a nerozli\u0161ujeme mezi sebou kuli\u010dky jedn\u00e9 barvy.<\/li>\n<li>Vyb\u00edr\u00e1me si karty z bal\u00ed\u010dku a nez\u00e1le\u017e\u00ed n\u00e1m na jejich po\u0159ad\u00ed, ale pouze na jejich barv\u011b \u010di typu karty.<\/li>\n<li>Nakupujeme 6 jogurt\u016f a m\u00e1me na v\u00fdb\u011br ze spousty vanilkov\u00fdch, \u010dokol\u00e1dov\u00fdch a nug\u00e1tov\u00fdch jogurt\u016f.<\/li>\n<\/ol>\n<ul>\n<li>Jak je z p\u0159\u00edklad\u016f patrn\u00e9, Kombinace s opakov\u00e1n\u00edm vyu\u017e\u00edv\u00e1me v\u017edy, kdy m\u00e1me na v\u00fdb\u011br z nep\u0159ebern\u00e9ho mno\u017estv\u00ed jogurt\u016f, kuli\u010dek, karet \u010di jin\u00fdch v\u011bc\u00ed. P\u0159i\u010dem\u017e n\u00e1m nez\u00e1le\u017e\u00ed na po\u0159ad\u00ed, v jak\u00e9m si dan\u00e9 v\u011bci vybereme, ale na po\u010dtu zp\u016fsob\u016f, kter\u00fdmi si dan\u00e9 v\u011bci m\u016f\u017eeme vybrat.<\/li>\n<\/ul>\n<p>Vzorov\u00fd p\u0159\u00edklad:<\/p>\n<p>Ur\u010dete po\u010det kv\u00e1dr\u016f, jejich\u017e velikosti hran jsou p\u0159irozen\u00e1 \u010d\u00edsla nejv\u00fd\u0161e rovn\u00e1 deseti. Kolik je v tomto po\u010dtu krychl\u00ed?<\/p>\n<p><strong>\u0158e\u0161en\u00ed:<\/strong><\/p>\n<p>Nejprve si mus\u00edme uv\u011bdomit co vlastn\u011b vyb\u00edr\u00e1me a z kolika mo\u017enost\u00ed.<\/p>\n<ul>\n<li>P\u0159irozen\u00e1 \u010d\u00edsla jsou \u010d\u00edsla od 1 do 10, celkem tedy hrany m\u016f\u017eeme vybrat z 10 mo\u017enost\u00ed \u21d2 <em><strong>n=10<\/strong><\/em><\/li>\n<li>Kv\u00e1dr je tvo\u0159en z hran <em><strong>a<\/strong><\/em>, <em><strong>b<\/strong><\/em> a<strong><em> c<\/em><\/strong> \u21d2 vyb\u00edr\u00e1me v\u017edy 3 \u010d\u00edsla \u21d2<strong><em> k=3<\/em><\/strong><\/li>\n<li>Nyn\u00ed ji\u017e m\u016f\u017eeme dosadit do vzorce a vypo\u010d\u00edtat:<\/li>\n<\/ul>\n<p>$$K\u00b4(k,n)=\\binom{n+k-1}{k}=\\binom{n+k-1}{n-1}=\\frac{(n+k-1)!}{k!(n-1)!}$$<\/p>\n<p>$$K\u00b4(k,n)=\\binom{10+3-1}{3}=\\binom{10+3-1}{10-1}=\\binom{12}{3}=\\frac{(10+3-1)!}{3!(n-1)!}=\\frac{12!}{3!9!}=220$$<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"comment_status":"closed","ping_status":"closed","template":"","class_list":["post-2595","lesson","type-lesson","status-publish","hentry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.1.1 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Kombinace s opakov\u00e1n\u00edm- z\u00e1kladn\u00ed vzorec + vzorov\u00fd p\u0159\u00edklad | Studijni-svet.cz<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/kombinace-s-opakovanim-zakladni-vzorec-vzorovy-priklad\/\" \/>\n<meta property=\"og:locale\" content=\"cs_CZ\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Kombinace s opakov\u00e1n\u00edm- z\u00e1kladn\u00ed vzorec + vzorov\u00fd p\u0159\u00edklad | Studijni-svet.cz\" \/>\n<meta property=\"og:description\" content=\"KOMBINACE S OPAKOV\u00c1N\u00cdM vyu\u017eijeme v p\u0159\u00edpad\u011b, \u017ee si vyb\u00edr\u00e1me prvky a nevad\u00ed n\u00e1m, \u017ee se n\u011bkter\u00e9 z nich budou opakovat. Z\u00e1kladn\u00ed vzorec pro v\u00fdpo\u010det Kombinac\u00ed s opakov\u00e1n\u00edm vypad\u00e1 takto: $$K\u00b4(k,n)=binom{n+k-1}{k}=binom{n+k-1}{n-1}=frac{(n+k-1)!}{k!(n-1)!}$$ P\u0159\u00edklady, kdy m\u016f\u017eeme vzorec pro Kombinace s opakov\u00e1n\u00edm vyu\u017e\u00edt, jsou nap\u0159\u00edklad: Vyb\u00edr\u00e1me si z velk\u00e9ho po\u010dtu barevn\u00fdch kuli\u010dek t\u0159eba 3, 4, 5 \u010di v\u00edce a ... 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Z\u00e1kladn\u00ed vzorec pro v\u00fdpo\u010det Kombinac\u00ed s opakov\u00e1n\u00edm vypad\u00e1 takto: $$K\u00b4(k,n)=binom{n+k-1}{k}=binom{n+k-1}{n-1}=frac{(n+k-1)!}{k!(n-1)!}$$ P\u0159\u00edklady, kdy m\u016f\u017eeme vzorec pro Kombinace s opakov\u00e1n\u00edm vyu\u017e\u00edt, jsou nap\u0159\u00edklad: Vyb\u00edr\u00e1me si z velk\u00e9ho po\u010dtu barevn\u00fdch kuli\u010dek t\u0159eba 3, 4, 5 \u010di v\u00edce a ... 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