Isibonelo Semibuzo Yengxoxo Ehlanganisiwe
Ukuhlanganiswa kuwumqondo obalulekile ku-probability theory kanye nezibalo, okusivumela ukubala inani lezindlela zokukhetha izinto kusuka kusethi ngaphandle kokubheka i-oda. Ekuhlanganisweni, i-oda lokukhetha alishintshi, ngokungafani ne-permutations, lapho i-oda lokukhetha liyisihluthulelo. Ukuhlanganiswa kusetshenziswa emikhakheni eyahlukahlukene, kusukela kwisayensi yekhompyutha kanye ne-biology kuya kwezomnotho kanye nezibalo. Lesi sihloko sizochaza umqondo oyisisekelo wokuhlanganiswa futhi sinikeze izibonelo eziningana zezinkinga kanye nezingxoxo ukusiza ukuqonda kwakho.
Umqondo Oyisisekelo Wokuhlanganiswa
Ngaphambi kokuthi siqhubekele ezinkingeni zesibonelo, ake siqale ngokuqonda okuyisisekelo kokuhlanganiswa.
Inhlanganisela yezinto ezithathwe ngesikhathi esisodwa iboniswa yinothi \(\binom{n}{r}\) noma \(\mathbf{C}(n, r)\). Ifomula yokubala inhlanganisela yile:
\[ \binom{n}{r} = \frac{n!}{r!(nr)!} \]
– \( n \) inani eliphelele lezinto ezitholakalayo.
– \( r \) inani lezinto okufanele zikhethwe.
– ! (i-factorial) umsebenzi wezibalo ophindaphinda inombolo ngazo zonke izinombolo ezinhle ngaphansi kwayo kuze kufike ku-1.
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1: Ukukhetha Ithimba Eqenjini Labafundi
Umbuzo:
Kunabafundi abayi-10 ekilasini. Zingaki izindlela abafundi abane ekilasini abangakhethwa ngazo ukuze bahlanganyele emncintiswaneni?
Ingxoxo:
Ukusebenzisa ifomula yokuhlanganisa:
\[ \binom{10}{4} = \frac{10!}{4! \cdot (10-4)!} = \frac{10!}{4! \cdot 6!} \]
Ukubala i-factorial:
– \( 10! = 10 \izikhathi ezingu-9 \izikhathi ezingu-8 \izikhathi ezingu-7 \izikhathi ezingu-6! \)
– \( 4! = 4 \izikhathi 3 \izikhathi 2 \izikhathi 1 = 24 \)
– \( 6! \) isivele iphezulu, ngakho-ke ingakhansela u-6! ku- \(10!\).
Ngakho-ke:
\[ \binom{10}{4} = \frac{10 \izikhathi ezingu-9 \izikhathi ezingu-8 \izikhathi ezingu-7}{4 \izikhathi ezingu-3 \izikhathi ezingu-2 \izikhathi ezingu-1} = \frac{5040}{24} = 210 \]
Ngakho-ke, kunezindlela ezingu-210 zokukhetha abafundi abangu-4 kubafundi abayi-10.
Isibonelo Umbuzo 2: Ukuhlanganiswa Kokusebenzisa Amasethi Amakhadi
Umbuzo:
Kusukela edekini elinamakhadi angu-52, zingaki izindlela zokukhetha isandla se-poker samakhadi ama-5?
Ingxoxo:
Ukusebenzisa ifomula yokuhlanganisa:
\[ \binom{52}{5} = \frac{52!}{5! \cdot (52-5)!} = \frac{52!}{5! \cdot 47!} \]
Ukubala i-factorial:
– \( 52! = 52 \izikhathi ezingu-51 \izikhathi ezingu-50 \izikhathi ezingu-49 \izikhathi ezingu-48 \izikhathi ezingu-47! \)
– \( 5! = 5 \izikhathi 4 \izikhathi 3 \izikhathi 2 \izikhathi 1 = 120 \)
– \( 47! \) isivele iphezulu, ngakho-ke ingakhansela u-47! ku- \(52!\).
Ngakho-ke:
\[ \binom{52}{5} = \frac{52 \izikhathi ezingu-51 \izikhathi ezingu-50 \izikhathi ezingu-49 \izikhathi ezingu-48}{5 \izikhathi ezingu-4 \izikhathi ezingu-3 \izikhathi ezingu-2 \izikhathi ezingu-1} = \frac{311875200}{120} = 2598960 \]
Ngakho-ke, kunezindlela ezingu-2,598,960 zokukhetha amakhadi angu-5 kumakhadi angu-52.
Isibonelo Umbuzo 3: Ukubala Izinhlanganisela Zokukhethwa Kweqembu Lezemidlalo
Umbuzo:
Iqembu lebhola lomnqakiswano linabadlali abayi-15. Uma umqeqeshi efuna ukukhetha abadlali aba-5 njengabaqalayo, zingaki izindlela lokhu okungenziwa ngazo?
Ingxoxo:
Ukusebenzisa ifomula yokuhlanganisa:
\[ \binom{15}{5} = \frac{15!}{5! \cdot (15-5)!} = \frac{15!}{5! \cdot 10!} \]
Ukubala i-factorial:
– \( 15! = 15 \izikhathi ezingu-14 \izikhathi ezingu-13 \izikhathi ezingu-12 \izikhathi ezingu-11 \izikhathi ezingu-10! \)
– \( 5! = 5 \izikhathi 4 \izikhathi 3 \izikhathi 2 \izikhathi 1 = 120 \)
– \( 10! \) isivele iphezulu, ngakho-ke ingakhansela u-10! ku- \(15!\).
Ngakho-ke:
\[ \binom{15}{5} = \frac{15 \izikhathi ezingu-14 \izikhathi ezingu-13 \izikhathi ezingu-12 \izikhathi ezingu-11}{5 \izikhathi ezingu-4 \izikhathi ezingu-3 \izikhathi ezingu-2 \izikhathi ezingu-1} = \frac{360360}{120} = 3003 \]
Ngakho-ke, kunezindlela ezingu-3,003 zokukhetha abadlali aba-5 kubadlali abayi-15.
Isibonelo Umbuzo 4: Ukuhlanganiswa Kwezinto Eziphilayo
Umbuzo:
Engadini, kunezinhlobo ezingu-8 ezahlukene zezimbali. Zingaki izindlela izinhlobo ezintathu zezimbali ezingakhethwa ngazo ukuze kwenziwe amalungiselelo ezimbali?
Ingxoxo:
Ukusebenzisa ifomula yokuhlanganisa:
\[ \binom{8}{3} = \frac{8!}{3! \cdot (8-3)!} = \frac{8!}{3! \cdot 5!} \]
Ukubala i-factorial:
– \( 8! = 8 \izikhathi 7 \izikhathi 6 \izikhathi 5! \)
– \( 3! = 3 \izikhathi 2 \izikhathi 1 = 6 \)
– \( 5! \) isivele iphezulu, ngakho-ke ingakhansela u-5! ku- \(8!\).
Ngakho-ke:
\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]
Ngakho-ke, kunezindlela ezingu-56 zokukhetha izinhlobo ezintathu zezimbali ezinhlotsheni ezingu-8 zezimbali.
Isibonelo Umbuzo 5: Ukuhlanganiswa Ekwakhiweni Kweqembu
Umbuzo:
Unezincwadi eziyi-12 ezahlukene eshalofini lezincwadi. Zingaki izindlela ongakhetha ngazo izincwadi ezi-5 kweziyi-12?
Ingxoxo:
Ukusebenzisa ifomula yokuhlanganisa:
\[ \binom{12}{5} = \frac{12!}{5! \cdot (12-5)!} = \frac{12!}{5! \cdot 7!} \]
Ukubala i-factorial:
– \( 12! = 12 \izikhathi ezingu-11 \izikhathi ezingu-10 \izikhathi ezingu-9 \izikhathi ezingu-8 \izikhathi ezingu-7! \)
– \( 5! = 5 \izikhathi 4 \izikhathi 3 \izikhathi 2 \izikhathi 1 = 120 \)
– \( 7! \) isivele iphezulu, ngakho-ke ingakhansela u-7! ku- \(12!\).
Ngakho-ke:
\[ \binom{12}{5} = \frac{12 \izikhathi ezingu-11 \izikhathi ezingu-10 \izikhathi ezingu-9 \izikhathi ezingu-8}{5 \izikhathi ezingu-4 \izikhathi ezingu-3 \izikhathi ezingu-2 \izikhathi ezingu-1} = \frac{95040}{120} = 792 \]
Ngakho-ke, kunezindlela ezingu-792 zokukhetha izincwadi ezi-5 ezincwadini ezingu-12.
Isiphetho
Ukuhlanganiswa kuwumqondo owusizo kakhulu emikhakheni eyahlukene, ikakhulukazi lapho sidinga ukwazi inani lezindlela zokukhetha izinto ngaphandle kokubheka ukuhleleka. Sisebenzisa ifomula eyisisekelo yokuhlanganiswa \(\binom{n}{r}\), singazibala ngokushesha nangokulula, uma nje siqonda ukuthi ama-factorial asebenza kanjani. Ngezibonelo ezingenhla, sithemba ukuthi sizoba nokuqonda okuphelele nokucacile kokuhlanganiswa. Qhubeka uzijwayeza ngezinkinga ezahlukahlukene ukuze uqinise ukuqonda kwakho.
