wai1994926 asked in 教育及參考書教學 · 10 years ago

# 中5級的數學題?5點

Mr Chan wants to invite 6 friends from a list of 9 friends to his party,find number of ways to select guest if

1. 2 of his friends cannot be invited together

2. 2 of his friends either be invited together or not be invited together

Rating
• 10 years ago

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