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標題:
maths nCr and nPr 2
發問:
please help me. I need steps. (solution at the bottom) thank you. 5 marks 2. (b) A school organized a table-tennis tournament. Any two players must play against each other. However, a player was quit after playing 3 matches and another player was absent from two matches, therefore these two players did not play... 顯示更多 please help me. I need steps. (solution at the bottom) thank you. 5 marks 2. (b) A school organized a table-tennis tournament. Any two players must play against each other. However, a player was quit after playing 3 matches and another player was absent from two matches, therefore these two players did not play against each other. If a total of 57 matches were played finally, how many players were there originally? Solution: 2. (b) 12
最佳解答:
Let n be the original no. of players, (n-1)C2+3-2+1=57 (n-1)(n-2)/2=55 n^2-3n+2=110 n^2-3n-108=0 (n-12)(n+9)=0 n=12 or 9 (rej.) Therefore, there are 12 players originally, Let A be the quit player, B be the player from 2 matches (n-1)C2 = total no. of matches played by ALL players including B but excluding A 3 = total no. of matches played by A -2 = total no. of matches B missed +1 because the the match that A and B did not play is counted twice (by +3 and -2)
其他解答:
maths nCr and nPr 2
發問:
please help me. I need steps. (solution at the bottom) thank you. 5 marks 2. (b) A school organized a table-tennis tournament. Any two players must play against each other. However, a player was quit after playing 3 matches and another player was absent from two matches, therefore these two players did not play... 顯示更多 please help me. I need steps. (solution at the bottom) thank you. 5 marks 2. (b) A school organized a table-tennis tournament. Any two players must play against each other. However, a player was quit after playing 3 matches and another player was absent from two matches, therefore these two players did not play against each other. If a total of 57 matches were played finally, how many players were there originally? Solution: 2. (b) 12
最佳解答:
Let n be the original no. of players, (n-1)C2+3-2+1=57 (n-1)(n-2)/2=55 n^2-3n+2=110 n^2-3n-108=0 (n-12)(n+9)=0 n=12 or 9 (rej.) Therefore, there are 12 players originally, Let A be the quit player, B be the player from 2 matches (n-1)C2 = total no. of matches played by ALL players including B but excluding A 3 = total no. of matches played by A -2 = total no. of matches B missed +1 because the the match that A and B did not play is counted twice (by +3 and -2)
其他解答:
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- s-t, v-t & a-t graph@1@
- (急) 用win XP燒o左隻碟然後Vista度開d相冇晒呀！！！點算？
- 我想要2009 第61屆校際音樂節報名表格，有冇人有﹖﹖( 20點!!!!!!! )@1@
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