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

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

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