(ii) Tus naj npawb ntawm kev ua haujlwm bijective f: [n] → [n] yog: n!=n(n−1)···(2)(1). (iii) Tus naj npawb ntawm kev txhaj tshuaj muaj peev xwm f: [k] → [n] yog: n(n−1)···(n−k+1). Pov thawj.
Koj pom tus lej ntawm cov haujlwm ua haujlwm li cas?
Expert Answer:
- Yog muaj nuj nqi txhais los ntawm teeb A rau teeb B f:A->B yog bijective, uas yog ib-ib thiab mus rau, ces n(A)=n(B)=n.
- Yog li thawj lub ntsiab ntawm teeb A tuaj yeem cuam tshuam nrog ib qho ntawm 'n' ntsiab hauv teeb B.
- Thaum thawj qhov cuam tshuam, qhov thib ob tuaj yeem cuam tshuam nrog ib qho ntawm qhov seem 'n-1' cov ntsiab lus hauv teeb B.
Yuav ua li cas muaj ntau yam bijective muaj?
Tam sim no nws tau muab tias hauv teeb A muaj 106 ntsiab. Yog li los ntawm cov ntaub ntawv saum toj no tus naj npawb ntawm bijective functions rau nws tus kheej (piv txwv li A rau A) yog 106!
Tus qauv ntawm tus lej yog dab tsi?
Yog tias lub teeb A muaj m ntsiab thiab teeb B muaj n ntsiab, ces tus lej ntawm kev ua tau los ntawm A mus rau B yog nm. Piv txwv li, yog teem A={3, 4, 5}, B={a, b}. Yog hais tias lub teeb A muaj m ntsiab thiab teeb B muaj n cov ntsiab lus, ces tus naj npawb ntawm kev ua haujlwm ntawm A mus rau B=nm - C1 (n-1)m + C2(n-2)m - C3(n-3)m+ …. - C -1 (1)m.
Koj pom pes tsawg lub luag haujlwm ntawm Arau B?
Tus naj npawb ntawm cov haujlwm ntawm A mus rau B yog |B|^|A|, lossis 32=9. Cia peb hais kom meej tias A yog qhov teeb {p, q, r, s, t, u}, thiab B yog ib qho teeb nrog 8 lub ntsiab lus txawv ntawm cov A. Cia peb sim los txhais cov haujlwm f: A → B. f(p) yog dab tsi?
