(2) 球號0,1,2,3之球個數相等(各5個)置於一箱,自其中任取8個球,
此8個球之球號總和除以4的餘數為0,1,2,3
試問這4種餘數的機率各是多少?
如何掌握任取k個球的變化??
算一個不如算一族
let F=(1+y)^5 * (1+xy)^5 *(1+xxy)^5 * (1+xxxy)^5
這是排列數的generating function
任取k個球,球號總和t 的組合數= x^t* y^k項係數
t=4m的排列數=N(總和餘數0)= [F(x=1)+F(x=i)+F(x=-1)+F(x=-i)]/4
=[(1+y)^20+(1-yyyy)^5+(1-yy)^10+(1-yyyy)^5]/4
N(任取8個球,總和餘數0)= y^8項係數=[C(20,8)+C(5,2)+C(10,4)+C(5,2)]/4
p(任取8個球,總和餘數0)=[C(20,8)+C(5,2)+C(10,4)+C(5,2)]/4C(20,8)
t=4m+2的排列數=N(總和餘數2)= [F(x=1)-F(x=i)+F(x=-1)-F(x=-i)]/4
=[(1+y)^20-(1-yyyy)^5+(1-yy)^10-(1-yyyy)^5]/4
N(任取8個球,總和餘數2)= y^8項係數=[C(20,8)-C(5,2)+C(10,4)-C(5,2)]/4
p(任取8個球,總和餘數2)=[C(20,8)-C(5,2)+C(10,4)-C(5,2)]/4C(20,8)
t=4m+1 (or 3)的排列數=N(總和餘數1)= [F(x=1)-iF(x=i)-F(x=-1)+iF(x=-i)]/4
=[(1+y)^20-(1-yy)^10]/4
N(任取8個球,總和餘數1)=n(任取8個球,總和餘數3)= y^8項係數=[C(20,8)-C(10,4)]/4
p(任取8個球,總和餘數1)= [C(20,8)-C(10,4)]/4C(20,8)
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當任取k=2n+1 個球:
t=4m的排列數=N(總和餘數0)= [F(x=1)+F(x=i)+F(x=-1)+F(x=-i)]/4
=[(1+y)^20+(1-yyyy)^5+(1-yy)^10+(1-yyyy)^5]/4
N(任取k個球,總和餘數0)= y^(2n+1)項係數=[C(20,k)+0+0+0]/4
p(任取k個球,總和餘數0)=1/4
t=4m+2的排列數=N(總和餘數2)= [F(x=1)-F(x=i)+F(x=-1)-F(x=-i)]/4
=[(1+y)^20-(1-yyyy)^5+(1-yy)^10-(1-yyyy)^5]/4
N(任取k個球,總和餘數2)= y^(2n+1)項係數=[C(20,8)-0+0-0]/4
p(任取k個球,總和餘數2)=1/4
t=4m+1 (or 3)的排列數=N(總和餘數1)= [F(x=1)-iF(x=i)-F(x=-1)+iF(x=-i)]/4
=[(1+y)^20-(1-yy)^10]/4
N(任取k個球,總和餘數1)=N(任取k個球,總和餘數3)= y^(2n+1)項係數=[C(20,k)-0]/4
p(任取8個球,總和餘數1)= 1/4
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當任取k=8n 個球:
t=4m的排列數=N(總和餘數0)= [F(x=1)+F(x=i)+F(x=-1)+F(x=-i)]/4
=[(1+y)^20+(1-yyyy)^5+(1-yy)^10+(1-yyyy)^5]/4
N(任取8n個球,總和餘數0)= y^8n項係數=[C(20,8n)+C(5,2n)+C(10,4n)+C(5,2n)]/4
p(任取8n個球,總和餘數0)=[C(20,8n)+C(5,2n)+C(10,4n)+C(5,2n)]/4C(20,8n)
t=4m+2的排列數=N(總和餘數2)= [F(x=1)-F(x=i)+F(x=-1)-F(x=-i)]/4
=[(1+y)^20-(1-yyyy)^5+(1-yy)^10-(1-yyyy)^5]/4
N(任取8n個球,總和餘數2)= y^8項係數=[C(20,8n)-C(5,2n)+C(10,4n)-C(5,2n)]/4
p(任取8n個球,總和餘數2)=[C(20,8n)-C(5,2n)+C(10,4n)-C(5,2n)]/4C(20,8n)
t=4m+1 (or 3)的排列數=N(總和餘數1)= [F(x=1)-iF(x=i)-F(x=-1)+iF(x=-i)]/4
=[(1+y)^20-(1-yy)^10]/4
N(任取8n個球,總和餘數1)=n(任取8n個球,總和餘數3)= y^8n項係數=[C(20,8n)-C(10,4n)]/4
p(任取8n個球,總和餘數1)= [C(20,8n)-C(10,4n)]/4C(20,8n)
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當任取k=8n-4 個球:
t=4m的排列數=N(總和餘數0)= [F(x=1)+F(x=i)+F(x=-1)+F(x=-i)]/4
=[(1+y)^20+(1-yyyy)^5+(1-yy)^10+(1-yyyy)^5]/4
N(任取k個球,總和餘數0)= y^k項係數=[C(20,k)-C(5,k/4)+C(10,k/2)-C(5,k/4)]/4
p(任取k個球,總和餘數0)=[C(20,k)-C(5,k/4)+C(10,k/2)-C(5,k/4)]/4C(20,k)
t=4m+2的排列數=N(總和餘數2)= [F(x=1)-F(x=i)+F(x=-1)-F(x=-i)]/4
=[(1+y)^20-(1-yyyy)^5+(1-yy)^10-(1-yyyy)^5]/4
N(任取k個球,總和餘數2)= y^k項係數=[C(20,k)+C(5,k/4)+C(10,k/2)+C(5,k/4)]/4
p(任取k個球,總和餘數2)=[C(20,k)+C(5,k/4)+C(10,k/2)+C(5,k/4)]/4C(20,k)
t=4m+1 (or 3)的排列數=N(總和餘數1)= [F(x=1)-iF(x=i)-F(x=-1)+iF(x=-i)]/4
=[(1+y)^20-(1-yy)^10]/4
N(任取k個球,總和餘數1)=n(任取k個球,總和餘數3)= y^k項係數=[C(20,k)-C(10,k/2)]/4
p(任取k個球,總和餘數1)= [C(20,k)-C(10,k/2)]/4C(20,k)
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當任取k=4n-2 個球:
t=4m的排列數=N(總和餘數0)= [F(x=1)+F(x=i)+F(x=-1)+F(x=-i)]/4
=[(1+y)^20+(1-yyyy)^5+(1-yy)^10+(1-yyyy)^5]/4
N(任取k個球,總和餘數0)= y^k項係數=[C(20,k)-C(10,k/2)]/4
p(任取k個球,總和餘數0)=[C(20,k)-C(10,k/2)]/4C(20,k)
t=4m+2的排列數=N(總和餘數2)= [F(x=1)-F(x=i)+F(x=-1)-F(x=-i)]/4
=[(1+y)^20-(1-yyyy)^5+(1-yy)^10-(1-yyyy)^5]/4
N(任取k個球,總和餘數2)= y^k項係數=[C(20,k)-C(10,k/2)]/4
p(任取k個球,總和餘數2)=[C(20,k)-C(10,k/2)]/4C(20,k)
t=4m+1 (or 3)的排列數=N(總和餘數1)= [F(x=1)-iF(x=i)-F(x=-1)+iF(x=-i)]/4
=[(1+y)^20-(1-yy)^10]/4
N(任取k個球,總和餘數1)=n(任取k個球,總和餘數3)= y^k項係數=[C(20,k)+C(10,k/2)]/4
p(任取k個球,總和餘數1)= [C(20,k)+C(10,k/2)]/4C(20,k)