In how many ways can 7 friends accomodate to 1 triple, 2 double hotel rooms. (Probability counting technique)?
回答 (2)
2016-03-02 8:15 am
• From the 7 friends, choose 3 (₇C₃) to accommodate the triple room.
• From the 4 friends left, choose 2 (₄C₂), and allow them to choose one of the double rooms (₂C₁).
• The 2 friends left take the double room left (1).
The number of way for accommodation
= ₇C₃ × ₄C₂ × ₂C₁ × 1
= 35 × 6 × 2 × 1
= 420
• From the 4 friends left, choose 2 (₄C₂), and allow them to choose one of the double rooms (₂C₁).
• The 2 friends left take the double room left (1).
The number of way for accommodation
= ₇C₃ × ₄C₂ × ₂C₁ × 1
= 35 × 6 × 2 × 1
= 420
2016-03-02 10:39 am
The triple can contain 7C3=35 sets,
Double room A can contain 2 of the remaining
4 in 4C2=6 pairs and the final pair goes in
double room B.
These give 35X6=210 ways.
Double room A can contain 2 of the remaining
4 in 4C2=6 pairs and the final pair goes in
double room B.
These give 35X6=210 ways.
收錄日期: 2021-04-18 14:32:29
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