Counting, combination problem

G

geonkyu

New member
Joined
Nov 18, 2019
Messages
3
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won
$10$
games and lost
$10$
games; there were no ties. How many sets of three teams
$\{A, B, C\}$
were there in which
$A$
beat
$B$
,
$B$
beat
$C$
, and
$C$
beat
$A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$



2016 amc 10B #22
 
geonkyu said:
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won
$10$
games and lost
$10$
games; there were no ties. How many sets of three teams
$\{A, B, C\}$
were there in which
$A$
beat
$B$
,
$B$
beat
$C$
, and
$C$
beat
$A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$



2016 amc 10B #22
Click to expand...
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.
 
Subhotosh Khan said:
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.
Click to expand...

There are
$21$
teams. Any of the
$\tbinom{21}3=1330$
sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
[asy]size(7cm);label(X,(5,5));label(Z,(10,0));label(Y,(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label(X,(20,5));label(Z,(25,0));label(Y,(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy]
But we know that every team beat exactly
$10$
other teams, so for each possible
$X$
at the head of a fork, there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$
. Therefore there are
$21\cdot\tbinom{10}2=945$
forks, and all the rest must be cycles.
Thus the answer is
$1330-945=385$
which is
$\boxed{\textbf{(A)}}$
.



i don't understand this part in the solution
"so for each possible
$X$
at the head of a fork, there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$
. Therefore there are
$21\cdot\tbinom{10}2=945$
forks"
 
geonkyu said:
There are
$21$
teams. Any of the
$\tbinom{21}3=1330$
sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
[asy]size(7cm);label(X,(5,5));label(Z,(10,0));label(Y,(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label(X,(20,5));label(Z,(25,0));label(Y,(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy]
But we know that every team beat exactly
$10$
other teams, so for each possible
$X$
at the head of a fork, there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$
. Therefore there are
$21\cdot\tbinom{10}2=945$
forks, and all the rest must be cycles.
Thus the answer is
$1330-945=385$
which is
$\boxed{\textbf{(A)}}$
.



i don't understand this part in the solution
"so for each possible
$X$
at the head of a fork, there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$
. Therefore there are
$21\cdot\tbinom{10}2=945$
forks"
Click to expand...
There are two sections to the part in question:

section 1:

there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$


section 2:

Therefore there are
$21\cdot\tbinom{10}2=945$
forks"

Do you understand either of those sections?
 
Subhotosh Khan said:
There are two sections to the part in question:

section 1:

there are always exactly
$\tbinom{10}2$
choices for
$Y$
and
$Z$


section 2:

Therefore there are
$21\cdot\tbinom{10}2=945$
forks"

Do you understand either of those sections?
Click to expand...

No, i don't understand at all about section 1 and 2
10C2 means choose 2 in 10 right? why should we do 10C2?
ok, i understand there are only 2 ways fork type and circle type.
but i don't understand how to find the number of fork type sets.
 
You must log in or register to reply here.
Top