傳球數論問題

一問題背景年156）8）二分組討論多向求解師丙乙乙丙1．甲丙生1甲乙丙乙丙甲.甲乙丙…1生22甲非11.2111甲非非非非甲112甲2212112111121121++2．生3將35.123456圖36兩33或434、5A2221244種1A22A22342435A222A2A1+A2.22生421用366三進一步探究師44生51n4生6nan02n種不同的傳球方法；②經過nb種，則有：①ann1=+2abn1nnnn1n1，anb=abnn1nn1b2a2aa0可且an11n1n1n1nn12n2n12n32a(a)a{1得333nn1n2n221)2nn(a即a=n1.333①nn11111112aaaaa,().{}即則n1n1nn1n1nn22222322323nnna1131232aaa2.1i）式可得a（nnn1nn1n1a1025.6的生7522nnana(a)bnP，an2nnnnnb1P(a)Pb)P1(a)0P(a)Pb)=1P(a)，+=a=b.n1n1n221nnnnnnn1nnnn111131312(a)(P(a))P(a)P{的323nnn1n1nn1111(naP(a)=()P(a)=P(a)n1n332332n12nnnnnnn22(2nna2P(a).n3nnn生833：n中16n11132種n1122圖4與異色。同色相當于合并，為aa。n1na32aaa2n1故=1即nnn1生8nn1生92n生甲乙丙丁127an3(01)(01)(01)13(31)(31)(31)23(61)(61)(61)343(601)(601)(601)5na3(a(a(a33即a；nnnnnn4n333(a1)(a1)(a1),n當an即.na4nnnnn3(1)3ann.4n生nn13a(na1nnn1anan1na.12nn1n233n當n33(1)3.an12n32a2n4n333n當n33(1)3a.an12n32n4n23(3nn.a4生32n33364a3A2342322第2313故333(34nna9.a.444n2a2當n時3(1)32kkknk1.na4kk1kk種方法。但事實上，第k次akk3(1)33(3k1k1kk3a3.akk44k1knk1n生N33m（mn(m1)(1)(m1)nna,nN.mna(m1)(a0)aa、nn1n17n生m1m21甲非非非非甲m1m2m21n2次n22N(mm球nn2；232nn1(n2n3C(m(mN1n32n4；3(n6)n324n4C(nC.兩相鄰的情況n種，故這時的情況數為2n32n4于是總數為NC((mm2n43n64n1n11nn((2)；Nmm222n4n2m(.2nnaN2nNN(mm2)n2n2nC(m(m2)C(m(m2)C(m(m2)1n32n42n43n63n54n8n12n1(m(m2，n(,mn2n3n2NC(m(m2)n2(k,k,(n奇或（n偶）.kk122kn(k2)(m((mnnnN,amn{N}生k生AA,A,,A123m另AAA1A1A1A23AAA122第n22AAAAmmmmA2AA第n3n次傳球1Amiin)A,A,,Ani12m,a,a,,aAa。1i2,i3,inA手中，所以第nAm,i,A,,A手123mA1aaa1，amaaam,n1,m2,n3,nm,nn2,n13,n1NN）4iinA,A,,A（m-2(3)若n12m3in1aaa2,i3,im,ia(mai2,i1aa(m2)a(2)2,ii12,i1(maa(ma（(m2)aai12,i22,i2,i22,i1aaa(maa)m1，（a2,2,1iia2,i2,i12,i12,i2a2,i12,i2a(aa)(m(m.a2nini2,i2,i12,n22,n15(maina又a）ni2,i2,i1m22,n1a(m2(m，a(m3(m(m，22,n22,n3a(m(m(m(m432,2,n4aa2,3(mk(mk(mk(k(m34542,n(n3)(m(m(m((m(k3k4k5k4n31[(m](m(n3n2n1((m(n4n31[(mmAAk11NA,A,,AAn223m1aaaaaaN2,23,2m,22,23,2m,2(ma(ma(m2)a](mma(m2)a]得N2,21,32,32,42,3(m(n3n2(mmaa)a](mmmn32,32,42,3m(m((mnn2.m(m((mnn2Nn.mm生8：m在n，n2傳n3m(m24中n個區(qū)域,2個用1(m((mnnn1A(ma)1有mm1nn域n121n1的涂色種數為A；②是區(qū)域n與區(qū)域1涂不同色，則有A種，故有n1nAAm(m,k,.k1kk1(A(Amm),k,n，1k1kkkk1nn(1)AA[(1)A(1)A]mm)m)m)k1k1n2nkn2kk1k3k3m(mA(m((m.由A得nnnA21nnB(m((mFAnmnna.1即mECn2D624__圖5圖664A(4((4732.1666234用m(m6中n1個圖7,n2mna有nmm13(m2)(1)(2)nnB[(m2)(1)(m2)].nnn6個41___4[(42)(1)(42)]120.4B555生正四面體的四個頂點記為1，2，3，4，從一點出發(fā)，等可能到其他3點，求從點1出發(fā)走7步又回到1的概率?？梢詫⑵渫茝V到一般情況：對于任意一個由m個點組成的網絡，如果對于這m個點中的任意一個點都與另外的m-1個點相連，那么從其中任意一個點