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mathmath333

  • one year ago

In a tournament each of the participants was to play one match against each of the other participants. 3 players fell ill after each of them had played 3 matches and had to leave the tournament . What was the total number of the participants at the beginning , if the total number of the matches played was 75 ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{In a tournament each of the participants was to play one match } \hspace{.33em}\\~\\ & \normalsize \text{against each of the other participants. 3 players fell ill after each} \hspace{.33em}\\~\\ & \normalsize \text{of them had played 3 matches and had to leave the tournament .} \hspace{.33em}\\~\\ & \normalsize \text{What was the total number of the participants at the beginning ,} \hspace{.33em}\\~\\ & \normalsize \text{if the total number of the matches played was 75 ?} \hspace{.33em}\\~\\ & a.)\ 8 \hspace{.33em}\\~\\ & b.)\ 10 \hspace{.33em}\\~\\ & c.)\ 12 \hspace{.33em}\\~\\ & d.)\ 15 \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    option d.) is given correct

  3. anonymous
    • one year ago
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    similar problem to Nourage's. look up the "handshake problem"

  4. imqwerty
    • one year ago
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    :D

  5. mathmath333
    • one year ago
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    that problem is easy this question has twists

  6. imqwerty
    • one year ago
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    ok what i think - let the no. of ppl be x number of matches played if we don't count those 3 ppl=(x-3)! nd if u count those 3 ppl we have to include the matches played by them =9 so (x-3)!+9+75 nd here comes up the twist

  7. imqwerty
    • one year ago
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    did i do any mistake out there? ^

  8. anonymous
    • one year ago
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    suppose there are 12 people in a room and each person is to shake the hand of everyone else in the room (excluding themself). the 1st person will shake hands with 11 others, the 2nd person will shake the hand of 10 others (already counted shaking hands with the 1st person), etc. 11+10+9+8+7+6+5+4+3+2+1 = 66 now 3 people shook hands with 3 people each for a total of 9. 66+9 = 75

  9. anonymous
    • one year ago
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    so, 15 people

  10. imqwerty
    • one year ago
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    wait i got my mistake :)

  11. anonymous
    • one year ago
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    \[\left(\begin{matrix}n \\ 2\end{matrix}\right)+9=75\]solve for n

  12. imqwerty
    • one year ago
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    it should be \[\frac{ (n-3)(2+(n-4)) }{ 2}\] instead of (n-3)! so we get- (n-3)(2+(n-4))/2 + 9 =75 n=14

  13. anonymous
    • one year ago
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    it's interesting... i put it all there and perhaps you don't understand but why offer something with no reasoning and the wrong answer to boot?

  14. anonymous
    • one year ago
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    \[\left(\begin{matrix}n \\ 2\end{matrix}\right)+9=75\]n=12 but you have to add 3 for the 3 that dropped out... n=12 is the number who stayed in and played each of the other 11 teams once/

  15. triciaal
    • one year ago
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    @UnkleRhaukus please help

  16. sohailiftikhar
    • one year ago
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    here we are given that each participant was to play one match against each one participant

  17. sohailiftikhar
    • one year ago
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    and let we consider participants 1,2,3,4,5,6,7etc we have to find the pairs

  18. sohailiftikhar
    • one year ago
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    like 1-2 ,1-3 ,1-4 ,1-5 etc but we are given that number of matches are 75 so we have to consider those numbers by which we can arrange that matches so start from 8

  19. sohailiftikhar
    • one year ago
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    hey got my point or not ?

  20. sohailiftikhar
    • one year ago
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    like we are given set {1,2,3,4,5,6} so we are asked that find the number of order pairs ..

  21. sohailiftikhar
    • one year ago
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    it looks like i'm talking to myself ..

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