Question

数学的排列组合问题

C94/96+C95/97+C2/98+C2/99=？C0/3+C1/4+C2/5+……+Ca/20=？

Answer 1

解:利用组合数性质:Cm/n=C(n-m)/n; Cm/(n+1)=Cm/n+C(m-1)/n;
C94/96+C95/97+C2/98+C2/99
=C(96-94)/96+C(97-95)/97+C2/98+C2/99
=C2/96+C2/97+C2/98+C2/99
=(C2/96+C3/96)-C3/96+C2/97+C2/98+C2/99
=C3/97+C2/97+C2/98+C2/99-C3/96
=C3/98+C2/98+C2/99-C3/96
=C3/99+C2/99-C3/96
=C3/100-C3/96
2.解:设n为项数,则C0/3+C1/4+C2/5+……+Ca/20=？式中每一项都是下标是项数+2,上标是项数-1,第n项是C(n-1)/(n+2)=n(n+1)(n+2)/3!=n(n+1)(n+2)/6.可以看出Ca/20是第18项,a=18.第1项C0/3=C0/4=1,所以有
C0/3+C1/4+C2/5+……+Ca/20
=C0/4+C1/4+C2/5+……+C18/20
=C1/5+C2/5+C3/6……+C18/20
=C2/5+C3/6……+C18/20 依此类推,由公式Cm/n+C(m-1)/n=Cm/(n+1)
=…
=C16/C19+C17/19+C18/20
=C17/20+C18/20
=C18/21
=C（21-18）/21
=C3/21
=1330

Answer 2

1.C94/96+C95/97+C2/98+C2/99
=C2/96+C2/97+C2/98+C2/99
=C3/97-C3/96+C2/97+C2/98+C2/99
=C3/98+C2/98+C2/99-C3/96
=C3/99+C2/99-C3/96
=C3/100-C3/96
2.C0/3+C1/4+.......+Ca/20
=C3/3+C3/4+C3/5+........C3/20
=C4/21=5985

Answer 3

C94/96+C95/97+C2/98+C2/99=96！/94！x2！+……99！/97！x2！=18820
C0/3+C1/4+C2/5+……+Ca/20=？这里面的a是什么，17吗，排列组合很早学的，忘得差不多了

Answer 4

C94/96+C95/97+C2/98+C2/99
=C2/96+C2/97+C2/98+C2/99
=(96*95+97*96+98*97+99*98)/2
=[96(95+97)+98(97+99)]/2
=96*96+98*98=18820.

C0/3+C1/4+C2/5+……+Ca/20 a=？
=