數學題(Combination)#10 15分

7(a) As Blue Jays needs to win until 7 games,

then the result of first six games should be 3-3

So, no. of ways = C(6,3) = 20

(b) As Expos needs to win until 6 games

then the result of first five games should be 3-2

So, no. of ways = C(5,3) = 10

(c) Consider Blue Jays

Wins in 4 games, 1 ways

Wins in 5 games, 4 ways

Wins in6 games, 10 ways (result of (b))

Wins in 7 games 20 ways (result of (a))

Total 35 ways

Similarly for Expos

Total possible ways = 2 * 35 = 70

10(a) 3 * 3 = 9 routes

(b) 3 * 2 = 6 routes (As a man cannot use the rural roads for return)

7a.
BJ should win the 7th game
so only first 6 games will be consider in the series
BJ should win 3 out of 6 games to get 7th game
the combination is 6C3 = 20
7b.
EX should win the 6th game
so only first 5 games will be consider in the series
EX should win 3 out of 5 games to get 7th game
the combination is 5C3 = 10
7c.
(consider BJ wins the series means BJ wins the lastmost game and so the games to be considered should be minus 1)
in 4 games = 3C0 = 1
in 5 games = 4C1 = 4
in 6 games = 5C2 = 10
in 7 games = 6C3 = 20
as there are 2 teams, so the total combination of series = 2x(1+4+10+20)
= 70

8. (what is the question?)

10a.
routes from A to B = 3
routes from B to A = 3
so total routes from A to B and back to A = 3x3 = 9
10b.
routes from B to A = 2
so total routes from A to B and back to A = 3x2 = 6

2011-12-16 22:59:53 補充：
7a.
the 7th game will play only when each team wins 3 games

7c.
e.g. BJ wins the series in 5 games
4 possible ways and you can realize BJ will get the last match:
EX-BJ-BJ-BJ-BJ
BJ-EX-BJ-BJ-BJ
BJ-BJ-EX-BJ-BJ
BJ-BJ-BJ-EX-BJ