can any one do number #6,10 & 11......thx very much every one......and show ur steps thnx.....cant thank you enough
[QUOTE=sonymd23]can any one do number #6,10 & 11......thx very much every one......and show ur steps thnx.....cant thank you enough
[qUOTE]
I'm pretty sure you will find the answer to 6 by hunting around this site for
what was said when these questions were previously posted.
RonL
When I look at this thread the old Tex does not render unless I edit aOriginally Posted by Quick
post (make no changes) then save. Is this what everyone else sees
happen ?
RonL
Looking at the printable version also works
This thread shows why I would prefer that the image file be uploaded here
rather than hosted somewhere like PhotoBucket - in that thread the questions
can no longer be seen as the file has been removed from PhotoBucket.
Hello, sonymd23!
however my teacher said thats incorrect......he wrote.....
John on floor: 8c2 x 7c2 x 5c5
" " to door : 8c1 x 8c3 x 5c5
" " as floater : 8c4 x 5c2 x 3c3
Are you sure this what he wrote?
Sorry, but I have no idea what your teacher is doing . . .
None of the expressions make sense . . .
He said "John on floor: $\displaystyle _8C_2 \times\, _6C_2 \times\, _5C_5$"
. . Does anyone else agree that this is totally contrary to any kind of logic?
It says: John is on the floor. .(Hence, Tom is not on the floor.)
Then $\displaystyle _8C_2$: we choose two from a group of eight people . . . for what?
. . Why pick two people? .What eight people?
If John is already assigned to the Floor and Tom is not, there are ten people to choose from.
Then $\displaystyle _6C_2$: we choose two from a group of six people.
. . Again, why two people . . . what six people?
Since John is already assigned and we've chosen 2 people (in the preceding),
. . then there are nine people to choose from (8, if we exclude Tom).
Finally $\displaystyle _5C_5$: take the remaining five people and place them anywhere?
Someone has been sipping way too much coughsyrup . . .
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The problem I had solved was different; it had ten people.
This problem has twelve people:
. . 2 at the door, 4 on the floor, 6 floaters.
Again, John and Tom will not work together.
I'll solve it both ways . . .
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There are three possible assignments for John.
[1] John is at the Door . . . and Tom is not.
The other one person at the Door is chosen from the other 10 people.
. . There are: $\displaystyle _{10}C_1 = 10$ ways.
The remaining 10 people are assigned to the 4 Floors and 6 Floaters.
. . There are $\displaystyle \begin{pmatrix}10\\4,6\end{pmatrix} = 210$ ways.
Hence, there are: $\displaystyle 10 \times 210 = \boxed{2100}$ ways that John can be at the Door.
[2] John is on the Floor . . . and Tom is not.
The other 3 people on the Floor are chosen from the other 10 people.
. . There are: $\displaystyle _}10}C_3 = 120$ ways.
The remaining 8 people are assigned to the 2 Doors and 6 Floaters.
. . There are: $\displaystyle \begin{pmatrix}8\\2,6\end{pmatrix} = 28$ ways.
Hence, there are: $\displaystyle 120 \times 28 = \boxed{3360}$ ways that John can be on the Floor.
[3] John is a Floater . . . and Tom is not.
The other 5 Floaters are chosen from the other 10 people.
. . There are: $\displaystyle _{10}C_5 = 252$ ways.
The remaing 5 people are assigned to the 2 Doors and 4 Floors.
. . There are: $\displaystyle \begin{pmatrix}6\\2,4\end{pmatrix} = 15$ ways.
Hence, there are: $\displaystyle 252 \times 15 = \boxed{3780}$ ways that John can be a Floater.
Therefore, there are: .$\displaystyle 2100 + 3360 + 3780 \:=$ 9240 ways to assign the positions.
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There are: $\displaystyle \begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860$ assignments with no restrictions.
Let's count the ways that John and Tom are together.
Duct-tape them together and we have 11 "people" to assign.
There are three cases . . .
$\displaystyle JT$ is at the Door.
Then there are $\displaystyle \begin{pmatrix}10\\4,6\end{pmatrix} = 210$ ways to assign the other ten people.
$\displaystyle JT$ is on the Floor.
Then there are $\displaystyle \begin{pmatrix}10\\2,2,6\end{pmatrix} = 1260$ ways to assign the other ten people.
$\displaystyle JT$ is a Floater.
Then there are $\displaystyle \begin{pmatrix}10\\2,4,4\end{pmatrix} = 3150$ ways to assign the other ten people.
Hence, there are $\displaystyle 210 + 1260 + 3150 = 4620$ ways for John and Tom to serve together.
Therefore, there are $\displaystyle 13,860 - 4620 \,=$ 9240 ways that John and Tom do not serve together.
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Who is right?
I'm through explaining . . . It's your call . . .
I thought the teacher's answer was for 10 volunteers...I am sorry.
Why did you use Hello Soroban!
$\displaystyle
\begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860
$
Please explain.
I thought it was used when some of the things are alike.
Keep Smiling
Malay
Hello, Malay!
Why did you use: $\displaystyle \begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860$
Please explain.
I thought it was used when some of the things are alike.
You are correct!
The twelve people are different (distinguisable), of course.
But selecting two people for the Door is a combination: $\displaystyle \begin{pmatrix}12\\2\end{pmatrix}$
. . The order in which the two people are chosen is not considered.
Then we select 4 people for the Floor from the remaining 10 people: $\displaystyle \begin{pmatrix}10\\4\end{pmatrix} $
. . Again, the order of the four names is not important.
Finally, we select 6 people to be Floaters from the remaining 6 people: $\displaystyle \begin{pmatrix}6\\6\end{pmatrix}$
. . And their order is ignored . . . so there is only one way.
The total number of assignments is: .$\displaystyle \begin{pmatrix}12\\2\end{pmatrix}\begin{pmatrix}10 \\4\end{pmatrix}\begin{pmatrix}6\\6\end{pmatrix}$
. . $\displaystyle = \;\frac{12!}{2!\not{10!}}\cdot\frac{\not{10!}}{4! \not{6!}}\cdot\frac{\not{6!}}{6!\not{0!}} \;=\;\frac{12!}{2!\,4!\,6!} \;= $ $\displaystyle \begin{pmatrix}12\\2,4,6\end{pmatrix}
$