五個人傳球

解答問題前, 先列出數個情況:

設 P(X) = X這種傳球方法的總數
例如X可以是 ABCD, 則代表A傳球給B, B傳球給C, C傳球給D.

P(M?N) = 3 ............. M不等如N
P(N?N) = 4
P(M??N) = 3xP(M?N) + P(N?N) = 3x3 + 4 = 13
P(N??N) = 4x3 = 12

***********************************************************************

(a) X = A???E

P(X)
= P(A???E)
= 3xP(M??E) + P(E??E)
= 3x13 + 12
= 51

參考:
BAB, BAC, BAD, BCA, BCB, BCD, BDA, BDB, BDC, BEA, BEB, BEC, BED, CAB, CAC, CAD, CBA, CBC, CBD, CDA, CDB, CDC, CEA, CEB, CEC, CED, DAB, DAC, DAD, DBA, DBC, DBD, DCA, DCB, DCD, DEA, DEB, DEC, DED, EAB, EAC, EAD, EBA, EBC, EBD, ECA, ECB, ECD, EDA, EDB, EDC

(b) X = A???(!C)?A ..................... ?(!C) 代表除C之外任何一個人
拆開前後來分析:

第一次至第三次傳球的情況:
P(A??A) = 12
P(A??B) = P(A??D) = P(A??E) = 13

第四次至第五次傳球的情況:
P(A?A) = 4
P(B?A) = P(D?A) = P(E?A) = 3

因此:
P(X)
= P(A??A?A) + P(A??B?A) + P(A??D?A) + (A??E?A)
= P(A??A) x P(A?A) + 3 x P(A??B) x P(B?A)
= 12x4 + 3x13x3
= 48 + 117
= 165

(c) X = A??(!C)?B?A

第一次至第二次傳球的情況:
P(A?A) = 4
P(A?B) = P(A?D) = P(A?E) = 3

第三次至第四次傳球的情況:
P(A?B) = P(D?B) = P(E?B) = 3
P(B?B) = 4

第一次至第四次傳球的情況:
P(A?A?B) = P(A?B?B) = P(A?A) x P(A?B) = 4x3 = 12
P(A?D?B) = P(A?E?B) = P(A?D) x P(D?B) = 3x3 = 9

第五次至第六次傳球的情況:
P(B?A) = 3

因此:
P(X)
= P(A?A?B?A) + P(A?B?B?A) + P(A?D?B?A) + P(A?E?B?A)
= (P(A?A?B) + P(A?B?B) + P(A?D?B) + P(A?E?B)) x P(B?A)
= (2x12 + 2x9) x 3
= 42 x 3
= 126

a)傳球方法最多有64種
b)傳球方法最多有192種
c)傳球方法最多有48種