Kombinatoryka, zadanie nr 6334
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
penelopa38 post贸w: 18 |  2020-05-20 09:40:57Bardzo prosz臋 o pomoc nie umiem tego rozwi膮za膰,prosz臋 o ca艂kowite rozwi膮zanie to Kombinacje z Newtonem zad1.Oblicz: ${8 \choose 2}$,${10 \choose 8}$,${12 \choose 10}$,${197 \choose 1}$,${37 \choose 36}$ zad.2 Zapisz w postaci iloczynu liczb a) ${24 \choose 20}$ b) ${100 \choose 97}$ c) ${13 \choose 5}$ d) ${26 \choose 13}$ zad.3 Wyznacz n a) ${n \choose 3}$-${n \choose 4}$=0 |
pm12 post贸w: 493 | 2020-05-22 10:58:17 1. ${8 \choose 2}$ = $\frac{8!}{2!\cdot(8-2)!}$ = $\frac{8!}{2!\cdot6!}$ = $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{(1\cdot2)\cdot(1\cdot2\cdot3\cdot4\cdot5\cdot6)}$ = $\frac{7\cdot8}{1\cdot2}$ = 28 |
pm12 post贸w: 493 | 2020-05-22 11:00:57 ${10 \choose 2}$ = $\frac{10!}{2!\cdot(10-2)!}$ = $\frac{10!}{2!\cdot8!}$ = $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}{(1\cdot2)\cdot(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8)}$ = $\frac{9\cdot10}{1\cdot2}$ = 45 |
pm12 post贸w: 493 | 2020-05-22 11:05:44 ${12 \choose 10}$ = $\frac{12!}{10!\cdot(12-10)!}$ = $\frac{12!}{10!\cdot2!}$ = $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12}{(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10)\cdot(1\cdot2)}$ = $\frac{11\cdot12}{1\cdot2}$ = 66 |
pm12 post贸w: 493 | 2020-05-22 11:09:25 ${197 \choose 1}$ = $\frac{197!}{1!\cdot(197-1)!}$ = $\frac{197!}{1!\cdot196!!}$ = $\frac{196!\cdot197}{1\cdot196!}$ = 197 |
pm12 post贸w: 493 | 2020-05-22 11:11:55 ${37 \choose 36}$ = $\frac{37!}{36!\cdot(37-36)!}$ = $\frac{37!}{36!\cdot1!}$ = $\frac{36!\cdot37}{36!\cdot1}$ = 37 |
pm12 post贸w: 493 | 2020-05-22 11:16:35 2. a) ${24 \choose 20}$ = $\frac{24!}{20!\cdot(24-20)!}$ = $\frac{24!}{20!\cdot4!}$ = $\frac{20!\cdot21\cdot22\cdot23\cdot24}{20!\cdot24}$ = $21\cdot22\cdot23$ = 10626 |
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