组合数学引论ch3答案.pdf

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1、

2、ÜêÆÚØ1nÙ‰Y2.x5y13Xê´µC55
(2)13=C55
213,18
318
3x8y10Xê´µC88
(2)10=C58
210.18
318
3!!!!nnnn3(2).ª†>=1[(n+1)+n+1+n+1+n+1+n+12340123!n


+n+1]n+1n!!!!n+1n+1n+1n+1=1[++


++]n+112nn+1=1(2n+11)n+1=m>!mm(m1)(m2)8.=;63!mm(m1)=;22!m=m;1!!!mmmm(m1)(m2)m(m1))a

3、+b+c=a+b+cm62321=am3+bam2+2a3b+6cm;62612)a=1;ba=0;2a3b+6c=0,626)a=6;b=6;c=1,13+23+


+n3!!!!!!111222=[(6+6+)+(6+6+)+


+321321!!!nnn(6+6+)]321!!!!!!12n12n=6(++


+)+6(++


+)+333222!!!12n


+(++


+)111!!!n+1n+1n+1=6+6+432!!n+2n+1=6+42n2(n+1)2=410.(1)y²µ•Ý•n

4、iÎG¥§0ÑyóêgiÎG‡ê•f(n)‡§ÑyÛêgiÎG‡ê•g(n)‡§@o§kf(n)=g(n1)+2f(n1)=3n1f(n1)+2f(n1)=3n1+f(n1),n1n213n13n+1f(n)=3+3+


+3+f(1)=(31)+2=22(2)ª†>Œ±n)••Ý••Ý•niÎG¥§0Ñyóê3giÎG‡êoÚ§=Ó1(1)¯"!!!n1n1n315.†>=+kk1k!!!!n2n2n1n3=++kk1k1k!!!!!n3n3n2n1n

5、3=+++kk1k1k1k=m>!!!Pm1m2m316.ª=r;s;t0r+s+t=n=nstst!!Pm1+m2m3=r;s;t0r+s+t=n=ntt!m1+m2+m3=n!nPnnk18.y²µ*(1+x)=k=0x;kªü>Óž¦,µ!n1Pnnk1n(1+x)=k=0kx;k!n1Pnnknx(1+x)=k=0kx;kü>2g¦§Óžx=1µ†>=n(n+1)2n2;4!Pn2nm>=k=1kk,)y..