# Math Help - 2 hard question....help plz!!!!

1. ## 2 hard question....help plz!!!!

can any one do number #6,10 & 11......thx very much every one......and show ur steps thnx.....cant thank you enough

2. [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

3. I know #10 was answered before by Soroban...

4. I think I found the post

5. Originally Posted by Quick
I think I found the post
When I look at this thread the old Tex does not render unless I edit a
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.

6. Originally Posted by CaptainBlack
When I look at this thread the old Tex does not render unless I edit a
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.
Yes, the LaTex is having some troubles, I wonder why? could it have anything to do with the fact you can click on it and get the code?

7. Try using a different internet brouswer.

8. but can any one do number 10?

9. Originally Posted by sonymd23
but can any one do number 10?
Originally Posted by MIA
10. A school dance has 10 volunteers. .Each dance requires 2 volunteers at the door,
3 volunteers on the floor, and 5 floaters. .If two of the volunteers, John and Tom,
can not work together, in how many ways can the volunteers be assigned?
Originally Posted by Soroban*
There are: $\binom{10}{2,3,5}\:=\:\frac{10!}{2!3!5!}\:=\;2520$ ways to make the assignments.

We will count the number of ways that John and Tom are together.
Duct-tape them together; then we have nine "people" to arrange.
. . $\{JT,\;A,\;B,\;C,\;D,\,E,\;F,\;G,\:H\}$

If JT are at the door, the other 8 people can be assigned in
. . $\binom{8}{3,5} \:=\:\frac{8!}{3!5!}\:=\:56 ways.$

If JT are on the floor, the other 8 people can be assigned in
. . $\binom{8}{2,1,5}\:=\:\frac{8!}{2!1!5!}\:=\:126 ways.$

If JT are floaters, the other 8 people can be assigned in
. . $\binom{8}{2,3,3}\:=\:\frac{8!}{2!3!3!}\:=\:560 ways.$

Hence, there are: $56 + 126 + 560\,=\,742$ ways that John and Tom are together.

Therefore, there are: $2520 - 742 \,=\,1778$ ways that they are not together.

10. Originally Posted by Quick

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

so which one is correct?

11. Originally Posted by 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

so which one is correct?
I think your teacher is correct.
I have one problem in Soroban,s solution: Why he had used the expression $\frac{10!}{5!3!2!}$?Please clarify.

Keep Smiling
Malay

12. 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: $_8C_2 \times\, _6C_2 \times\, _5C_5$"
. . 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 $_8C_2$: 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 $_6C_2$: 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 $_5C_5$: take the remaining five people and place them anywhere?

Someone has been sipping way too much coughsyrup . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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 . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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: $_{10}C_1 = 10$ ways.
The remaining 10 people are assigned to the 4 Floors and 6 Floaters.
. . There are $\begin{pmatrix}10\\4,6\end{pmatrix} = 210$ ways.
Hence, there are: $10 \times 210 = \boxed{2100}$ 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: $_}10}C_3 = 120$ ways.
The remaining 8 people are assigned to the 2 Doors and 6 Floaters.
. . There are: $\begin{pmatrix}8\\2,6\end{pmatrix} = 28$ ways.
Hence, there are: $120 \times 28 = \boxed{3360}$ 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: $_{10}C_5 = 252$ ways.
The remaing 5 people are assigned to the 2 Doors and 4 Floors.
. . There are: $\begin{pmatrix}6\\2,4\end{pmatrix} = 15$ ways.
Hence, there are: $252 \times 15 = \boxed{3780}$ ways that John can be a Floater.

Therefore, there are: . $2100 + 3360 + 3780 \:=$ 9240 ways to assign the positions.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

There are: $\begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860$ 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 . . .

$JT$ is at the Door.
Then there are $\begin{pmatrix}10\\4,6\end{pmatrix} = 210$ ways to assign the other ten people.

$JT$ is on the Floor.
Then there are $\begin{pmatrix}10\\2,2,6\end{pmatrix} = 1260$ ways to assign the other ten people.

$JT$ is a Floater.
Then there are $\begin{pmatrix}10\\2,4,4\end{pmatrix} = 3150$ ways to assign the other ten people.

Hence, there are $210 + 1260 + 3150 = 4620$ ways for John and Tom to serve together.

Therefore, there are $13,860 - 4620 \,=$ 9240 ways that John and Tom do not serve together.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Who is right?
I'm through explaining . . . It's your call . . .

13. I thought the teacher's answer was for 10 volunteers...I am sorry.

Why did you use Hello Soroban!
$
\begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860
$

I thought it was used when some of the things are alike.

Keep Smiling
Malay

14. Hello, Malay!

Why did you use: $\begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860$

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: $\begin{pmatrix}12\\2\end{pmatrix}$
. . 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: $\begin{pmatrix}10\\4\end{pmatrix}$
. . Again, the order of the four names is not important.

Finally, we select 6 people to be Floaters from the remaining 6 people: $\begin{pmatrix}6\\6\end{pmatrix}$
. . And their order is ignored . . . so there is only one way.

The total number of assignments is: . $\begin{pmatrix}12\\2\end{pmatrix}\begin{pmatrix}10 \\4\end{pmatrix}\begin{pmatrix}6\\6\end{pmatrix}$

. . $= \;\frac{12!}{2!\not{10!}}\cdot\frac{\not{10!}}{4! \not{6!}}\cdot\frac{\not{6!}}{6!\not{0!}} \;=\;\frac{12!}{2!\,4!\,6!} \;=$ $\begin{pmatrix}12\\2,4,6\end{pmatrix}

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