Imibuzo Eyisibonelo Exoxa Ngemithetho Yokugcwalisa Izikhala
Umthetho wokugcwalisa indawo, noma umthetho wokubeka, ungumqondo oyisisekelo kwizibalo kanye namathuba awusizo kakhulu ezimweni eziningi. Lo mthetho uvame ukusetshenziswa kumongo wokuhlela izinto ngokulandelana okuthile noma ekuhleleni okuhlukile. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zezinkinga ezihilela umthetho wokugcwalisa indawo, sinikeze izixazululo ezinemininingwane ngayinye.
I-Pendahuluan
Ukugcwalisa isikhala kuyindlela evamile esetshenziswa ku-combinatorics, insimu yezibalo efunda ukuhlelwa, ukuhlanganiswa, kanye nokukhethwa kwezinto. Esinye sezimiso eziyisisekelo ze-combinatorics umthetho wokuphindaphinda, othi uma kunezigaba eziningana enkambisweni futhi isigaba ngasinye sinenani elithile lokukhetha, khona-ke inani eliphelele lokuhlelwa okungenzeka lingatholakala ngokuphindaphinda inani lokukhetha esigabeni ngasinye.
Isibonelo, uma sinezigaba ezimbili lapho isigaba sokuqala sinezinketho ze-\(m\) kanye nesigaba sesibili sinezinketho ze-\(n\), khona-ke inani eliphelele lamalungiselelo angaba khona yi-\(m \times n\).
Ake sisebenzise lo mqondo ukuxazulula ezinye zezinkinga eziyisibonelo.
Isibonelo 1: Ukuhlela Izincwadi Eshalofini
Umbuzo:
Kunezincwadi ezi-5 ezahlukene kanye neshelufu lezincwadi elinezikhala ezi-5 zokugcwalisa. Zingaki izindlela izincwadi ezinhlanu ezingahlelwa ngazo eshelufini?
Ingxoxo:
Kulesi simo, sidinga ukuhlela izincwadi ezinhlanu ezindaweni ezinhlanu ezahlukene. Lena inkinga yokuguquguquka kwesimo ngoba ukuhleleka kubalulekile. Singasebenzisa umthetho wokugcwalisa isikhala noma umthetho wokuphindaphinda ukuxazulula le nkinga.
1. Ekamelweni lokuqala, sinezinketho ezi-5 zezincwadi.
2. Ngemva kokubekwa kwencwadi eyodwa ekamelweni lokuqala, sisele nezinketho ezine zezincwadi zegumbi lesibili.
3. Egumbini lesithathu, sinezinketho ezintathu ezisele zezincwadi, njalo njalo.
Isibalo senani eliphelele lezilungiselelo yilesi:
\[ 5 \izikhathi 4 \izikhathi 3 \izikhathi 2 \izikhathi 1 = 5! = 120 \]
Ngakho-ke, kunezindlela ezingu-120 zokuhlela izincwadi ezinhlanu.
Isibonelo sesi-2: Ukwakha Amagama Ngezinhlamvu Ezihlukene
Umbuzo:
Mangaki amagama ahlukene angakhiwa kusetshenziswa zonke izinhlamvu ezisegameni elithi “IZIBALO”, ngaphandle kokuziphinda?
Ingxoxo:
Okokuqala sidinga ukubona ukuthi zingaki izinhlamvu ezisegameni elithi "IZIBALO." Kunezinhlamvu ezingu-11, ezinye zazo eziphindaphindwayo. Izinhlamvu eziphindaphindwayo yilezi:
– M abaningi kuze kufike ku-2
- A kuze kufike ku-3
– T abaningi kuze kufike ku-2
– Ezinye izinhlamvu (E, I, K) zivela kanye.
Sisebenzisa ifomula yokuguqulwa kwezinto eziphindaphindwayo, okungukuthi:
\[ \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \]
lapho \( n \) kuyinani eliphelele lezinto (izinhlamvu) kanye \( n_1, n_2, \ldots, n_k \) kuyinani lokuphindaphinda kwento ngayinye ehlukile.
Ngegama elithi “IZIBALO”:
\[ n = 11, n_1 = 2 \umbhalo{ (M)}, n_2 = 3 \umbhalo{ (A)}, n_3 = 2 \umbhalo{ (T)}, n_4 = 1 \umbhalo{ (E)}, n_5 = 1 \umbhalo{ (I)}, n_6 = 1 \umbhalo{ (K)} \]
Ngakho inani lamagama angakhiwa yileli:
\[ \frac{11!}{2! \izikhathi 3! \izikhathi 2! \izikhathi 1! \izikhathi 1! \izikhathi 1!} = \frac{39916800}{2 \izikhathi 6 \izikhathi 2 \izikhathi 1 \izikhathi 1 \izikhathi 1} = \frac{39916800}{24} = 1663200 \]
Kunamagama ahlukene angu-1,663,200 angakhiwa.
Isibonelo 3: Ukunquma Inani Lenhlanganisela eMartabak
Umbuzo:
Umthengisi we-martabak unikeza izinketho ezinhlanu zokugcwalisa (ushizi, ushokoledi, amantongomane, ubhanana, nama-raisin). Uma ikhasimende lifuna ukukhetha ezintathu kwezinhlanu zokugcwalisa ze-martabak yalo, zingaki izinhlanganisela ezahlukene elingakhetha kuzo?
Ingxoxo:
Lena inkinga yokuhlanganiswa, hhayi ukuguquguquka, ngoba ukuhleleka akubalulekile. Sisebenzisa ifomula yokuhlanganiswa:
\[ C(n, k) = \frac{n!}{k!(nk)!} \]
lapho \( n \) kuyinani eliphelele lezinketho, kanye \( k \) kuyinani lezinketho ezithathiwe.
Kulokhu, \( n = 5 \) kanye \( k = 3 \), ngakho-ke:
\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \izikhathi 2!} = \frac{120}{6 \izikhathi 2} = \frac{120}{12} = 10 \]
Kunezinhlobo eziyi-10 ezahlukene zokukhetha okuqukethwe okungu-3 kusuka kuzinketho ezi-5.
Isibonelo 4: Ukuhlelwa Kwabahlanganyeli Emncintiswaneni
Umbuzo:
Kunabahlanganyeli abangu-8 emjahweni wokugijima. Zingaki izindlela abaqedi abathathu abaphezulu abangabekwa ngazo?
Ingxoxo:
Lena inkinga yokuguquguquka ngaphandle kokuphindaphinda ngoba isikhundla sisho ukuthi ukuhleleka kubalulekile. Sisebenzisa ifomula yokuguquguquka:
\[ P(n, k) = \frac{n!}{(nk)!} \]
Kulokhu, \( n = 8 \) kanye \( k = 3 \), bese kuba:
\[ P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{40320}{120} = 336 \]
Ngakho-ke, kunezindlela ezingu-336 zokubeka izikhundla ezintathu eziphezulu zabahlanganyeli abangu-8.
Kulesi sihloko, sixoxe ngezinkinga eziningana zezibonelo kanye nezixazululo zazo sisebenzisa imithetho yokugcwalisa isikhala ezimweni ezahlukahlukene: kusukela ekuhleleni izincwadi eshalofini kuya ekunqumeni ophumelele emncintiswaneni. Ukuqonda lezi zisekelo kuzokunika ukuzethemba okwengeziwe ekuxazululeni izinkinga ezahlukahlukene zokuhlanganisa kanye namathuba ongase uhlangane nawo.
