How many ways could a class of 18 students divide into groups of 3 students each?
Have 18 students take 3 out, then do that 6 times?
6(18P3)
is that right?
Thanks
How many ways could a class of 18 students divide into groups of 3 students each?
Have 18 students take 3 out, then do that 6 times?
6(18P3)
is that right?
Thanks
Try [tex]\L\frac{{18!}}{{\left( {6!} \right)\left( {3!} \right)^6 }}[/tex]
Can you explain why that works?
EDIT: I first read it as three groups not three in a group.
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Hello, wind!
How many ways could a class of 18 students divide into groups of 3 students each?
I'm sure that pka meant: [tex]\L\:\frac{18!}{6!(3!)^6}[/tex]
Let's baby-step through this problem . . .
First, we'll number the groups, 1 to 6.
For group #1, there are: [tex]\,{18 \choose 3}[/tex] ways.
For group #2, there are: [tex]\,{15 \choose 3}[/tex] ways.
For group #3, there are: [tex]\,{12 \choose 3}[/tex] ways.
For group #4, there are: [tex]\,{9 \choose 3}[/tex] ways.
For group #5, there are: [tex]\,{6 \choose 3}[/tex] ways.
For group #6, there are: [tex]\,{3 \choose 3}[/tex] ways.
Hence, there are: [tex]\L\:\frac{18!}{3!\sout{15!}}\cdot\frac{\sout{15!}} {3!\sout{12!}}\cdot\frac{\sout{12!}}{3!\sout{9!}}\ cdot\frac{\sout{9!}}{3!\sout{6!}}\cdot\frac{\sout{ 6!}}{3!\sout{3!}}\cdot\frac{\sout{3!}}{3!\sout{0!} } \:=\:\frac{18!}{(3!)^6}[/tex] ways.
But the groups are not numbered; their order is not considered.
Since our answer includes the [tex]6![/tex] different orders of the six groups,
. . we must divide by [tex]6![/tex]
Answer: [tex]\L\:\frac{18!}{6!(3!)^6}[/tex]
I'm the other of the two guys who "do" homework.
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