mathmath333
  • mathmath333
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{In a tournament each of the participants was to play one match } \hspace{.33em}\\~\\ & \normalsize \text{against each of the other participants. 3 players fell ill after each} \hspace{.33em}\\~\\ & \normalsize \text{of them had played 3 matches and had to leave the tournament .} \hspace{.33em}\\~\\ & \normalsize \text{What was the total number of the participants at the beginning ,} \hspace{.33em}\\~\\ & \normalsize \text{if the total number of the matches played was 75 ?} \hspace{.33em}\\~\\ & a.)\ 8 \hspace{.33em}\\~\\ & b.)\ 10 \hspace{.33em}\\~\\ & c.)\ 12 \hspace{.33em}\\~\\ & d.)\ 15 \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
option d.) is given correct
anonymous
  • anonymous
similar problem to Nourage's. look up the "handshake problem"

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imqwerty
  • imqwerty
:D
mathmath333
  • mathmath333
that problem is easy this question has twists
imqwerty
  • imqwerty
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
imqwerty
  • imqwerty
did i do any mistake out there? ^
anonymous
  • anonymous
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
anonymous
  • anonymous
so, 15 people
imqwerty
  • imqwerty
wait i got my mistake :)
anonymous
  • anonymous
\[\left(\begin{matrix}n \\ 2\end{matrix}\right)+9=75\]solve for n
imqwerty
  • imqwerty
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
\[\left(\begin{matrix}n \\ 2\end{matrix}\right)+9=75\]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/
triciaal
  • triciaal
@UnkleRhaukus please help
sohailiftikhar
  • sohailiftikhar
here we are given that each participant was to play one match against each one participant
sohailiftikhar
  • sohailiftikhar
and let we consider participants 1,2,3,4,5,6,7etc we have to find the pairs
sohailiftikhar
  • sohailiftikhar
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
sohailiftikhar
  • sohailiftikhar
hey got my point or not ?
sohailiftikhar
  • sohailiftikhar
like we are given set {1,2,3,4,5,6} so we are asked that find the number of order pairs ..
sohailiftikhar
  • sohailiftikhar
it looks like i'm talking to myself ..

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