中5級的數學題?5點

1) Method 1 :All ways - 2 of his friends be invited together ways= 9C6 - (2C2) * (9-2)C(6-2)
= 9C6 - 7C4
= 84 - 35
= 49 ways.
Method 2 :None of the 2 friends be invited ways + only 1 out of the 2 friends be invited ways
= (9-2)C6 + (2C1) (9-2)C(6-1)
= 7C6 + 2 * 7C5
= 7 + 2 * 21
= 49 ways
Method 3 :At most 1 out of the 2 friends be invited ways
= Only 1 out of the 2 friends be invited ways - None of the 2 friends be invited ways
= (2C1) * (9-1)C6 - (9-2)C6
= 2 * 8C6 - 7C6
= 2 * 28 - 7
= 49 ways
2)Method 1 : Be invited together ways + not be invited together ways
= All ways
= 9C6
= 84 ways
Method 2 :Be invited together ways + not be invited together ways
= Be invited together ways + At most 1 out of the 2 friends be invited ways
= (2C2) * (9-2)C(6-2) + (2C1) * (9-1)C6 - (9-2)C6
= 35 + 49
= 84 ways

2011-05-10 16:55:20 補充：
Corrections of Q2) :

Method 1 :

be invited together ways + not be invited together ways

= (2C2) * (9-2)C(6-2) + (9-2)C6

= 35 + 7

= 42 ways

2011-05-10 16:55:26 補充：
Method 2 :

be invited together ways + not be invited together ways

= All ways - only 1 out of the 2 friends be invited ways

= 9C6 - (2C1) (9-2)C(6-1)

= 84 - 42

= 42 ways

2011-05-10 16:56:17 補充：
Corrections of Q2) :

Method 1 :

be invited together ways + not be invited together ways

= (2C2) * (9-2)C(6-2) + (9-2)C6

= 35 + 7

= 42 ways

2011-05-10 16:56:32 補充：
Method 2 :

be invited together ways + not be invited together ways

= All ways - only 1 out of the 2 friends be invited ways

= 9C6 - (2C1) (9-2)C(6-1)

= 84 - 42

= 42 ways