Tsarin haɗin kai shine: ncr = n! ((n u2013 r)! ku r!) n = adadin abubuwa.
Anan, Ta yaya kuke lissafin misalin haɗin gwiwa? Ana amfani da dabarar haɗin gwiwa don nemo adadin hanyoyin zabar abubuwa daga tarin, kamar yadda tsarin zaɓin ba shi da mahimmanci.
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Formula don Haɗuwa.
Tsarin Haɗuwa | nCr=n!(nu2212r)!r! ncr = n! ( n u2212 r ) ! r! |
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Tsarin Haɗuwa ta Amfani da Permutation | C (n, r) = P(n,r)/ r! |
Menene hade da misali? Haɗin kai zaɓi ne na duka ko ɓangarorin saitin abubuwa, ba tare da la'akari da tsarin da aka zaɓa ba. Misali, a ce muna da saitin haruffa uku: A, B, da C.… Kowane zaɓi mai yuwuwa zai kasance misali na hadewa. Cikakkun jerin zaɓuɓɓukan da za a iya yi zasu kasance: AB, AC, da BC.
Bugu da ƙari, wace hanya ce mafi sauƙi don lissafin haɗuwa?
Menene darajar 8C5? (n-r)! 8C5=8!
Menene darajar 5c 2?
5 ZABE 2 = 10 haɗuwa mai yiwuwa. 10 shine jimillar duk haɗin haɗin kai don zaɓar abubuwa 2 a lokaci ɗaya daga abubuwa daban-daban 5 ba tare da la'akari da tsari na abubuwa a cikin ƙididdiga & binciken yiwuwar ko gwaje-gwaje ba.
Menene darajar haɗin 8 5? (n-r)! = (8-5)! (8-5)! = 3!
Menene darajar 10C 3? C3= 10 ! / 3 ! (7) ku!
Menene darajar 6C4?
(n-r)! r ! 6C4=6!
Hakanan Menene darajar 7v4? Taƙaitaccen: Ƙa'idar ko haɗin kai 7C4 is 35.
Menene amsar 5C3?
Combinatorics da Pascal's Triangle
0C0 = 1 | ||
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2C0 = 1 | 2C1 = 2 | |
3C0 = 1 | 3C2 = 3 | |
4C0 = 1 | 4C1 = 4 | 4C2 = 6 |
5C1 = 5 | 5C3 ku 10 |
Menene ma'anar 3C2? 3v2. =3! (2!) (3-2)! =3!
Menene darajar 10C 4?
Bayanin mataki-mataki:
10 zabi 4 = 201 haɗuwa mai yiwuwa. 201 shine jimillar duk haɗin haɗin kai don zaɓar abubuwa 4 a lokaci ɗaya daga zuwa abubuwa daban ba tare da la'akari da tsari na abubuwa a cikin ƙididdiga & binciken yiwuwar ko gwaji ba.
Menene darajar 6C 2?
Nemo 6C2. 6C2 = 6!/(6-2)! 2 ! = 6 ! / 4!
Haɗin kai nawa ne na lambobin 1 2 3 4 akwai? Bayani: Idan muna duban adadin lambobin da za mu iya ƙirƙira ta amfani da lambobi 1, 2, 3, da 4, za mu iya ƙididdige hakan ta hanyar haka: ga kowane lambobi (dubu, ɗaruruwa, goma, goma, ɗaya), muna da 4. zabin lambobi. Don haka za mu iya ƙirƙirar 4×4×4×4=44=Lambobi 256.
Ta yaya kuke warware Factorials guda 10? ya kai 362,880. Gwada kirga guda 10! 10! = 10×9!
Menene 4C1?
4 ZABI 1 = 4 yuwuwar haɗuwa. Bayani: Yanzu yadda yake faruwa Don haka, 4 shine jimillar duk haɗin haɗin gwiwa don zaɓar abubuwa 1 a lokaci ɗaya daga abubuwa daban-daban guda 4 ba tare da la'akari da tsari na abubuwa a cikin ƙididdiga & binciken yiwuwar ko gwaje-gwaje ba. Godiya 0.
Menene darajar 5C1? Combinatorics da Pascal's Triangle
2C0 = 1 | 2C2 = 1 | |
3C0 = 1 | 3C2 = 3 | |
4C0 = 1 | 4C1 = 4 | 4C3 = 4 |
5C1 ku 5 | 5C3 = 10 |
Menene darajar 6P4?
⇒6P4=6! (6-4)! =6!
Menene haɗin 15c3? 0
Menene haɗin 4C2?
Mun san cewa dabarar da aka yi amfani da ita don magance maganganun haɗin gwiwa an ba da ita ta: …Masanya n = 4 da r = 2 a cikin dabarar da ke sama, 4C2 = 4! 2! (4-2) = 4! (2!
Menene 7c3? 8×7×6=336. C7,3=7!( 3!) ( 7-3)! = 7!
Ta yaya kuke warware 5P2?
5P2 = ku 5! (5-2)! = 5x4x3! / 3!
Yaya kuke yin 5C3 akan kalkuleta?
Menene 10C7?
⇒10C7=10! 7! ×3! =10×9×8×7×6×5×4×3×2 7×6×5×4×3×2 ×3×2. =10×9×83×2=120.
Menene haɗin 5C4?
nCr=(r!)(n-r)! ba! Saboda haka, 5C4=(4!)(