Math HW help please!!

[desi43]

Hi! Was was just wondering if someone can help me with my math problems? I have gotten answers to most of the questions except for a few, or parts of a few. so I would really appreciate if you could check my work on the ones I did and help me with the two I couldn't figure out??

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1.

a) how many diffrent ways can 6 students stand in a row?

6x6x6x6x6x6= 46,656

b) if two of the students (john & cindy) wish to stand next to eachother?

i am really not sure on this one

1x1x4x4x4x4= 265

c) if two students refuse to stand next to eacother?

i can't figure this part out


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2. how many diffrent ways can 5 students be arranged in a circle?

5!=120 ways

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3.
an employee association allows it's members to vote yes, no or abstain on any given ballot proposition.

a) if a group has 8 members, how many diffrent results are possible.

i think this might be

3x3x3x3x3x3x3x3= 6561

b) if a proposition needs 5 (or more) yes votes to pass? how many ways can that be done?

c(8,8) =1
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thank you so much!

 

[stapel]

1-a) I'm not sure how you're getting that each position has six options. After you've picked the first person in line (there would be six ways to do this), wouldn't you have only five choices left for the second person in line?

1-b) If two of the students are to be regarded as a unit, then you have five positions to fill, after a fashion. Once you've filled those five positions (in the same manner as in (a) above), then multiply by "2", to account for the fact that you don't know which of the two students is in front of the other.

1-c) Pick one of those students to be first in line. There's two ways to do that. Since one of the five remaining students will not take the second place, how many ways can you chose somebody for the second position? Now that you've got somebody between the two who don't like each other, how many ways can you fill the third position? And so forth.

But... I'm not sure how to account for the fact that one of those students doesn't have to be first in line....

2) You're on the right track, but I think you omitted something from the formula...?

3-a) I think you're correct.

3-b) Since you know that YYYYYNNN, YYYYYYNN, YYYYYYYN, and YYYYYYYY are among the valid wasy to pass the proposition, "one way" seems unlikely to be the correct solution.

Eliz.

 

[Gene]

Note: Use * for times, not x.
1a. Not quite. After the first stands in the first spot there are only 5 for the second spot, 4 for the third...
6*5*4*3*2*1 = 6! =...

1b. Think of them as one person. Now there are 5 the first spot, 4 for the second spot...
5*4*3*2*1
but john can be first or second so
5*4*3*2*1*2 = 5!*2

1c. line up the four with 5 empty spaces.
_4_3_2_1_
There are 5 placess for the first and 4 for the second, then close them up.
4*3*2*1*5*4 = ...

2. Close but in a circle the first starts the counting so there is just 1 place for him. Then it is 4! for the rest.
1*4! = ...

3a. But what is a result?
With one person in the group
Y,N,A = 3 results
With two people in the group
YY, YN, YA
NY, NN, NA
AY, AN, AA
But YN is the same result as NY. one yes, one no
YY, YN, YA
__, NN, NA
__, __, AA
6 results
Would youlike to try this again?

3b. Are you familiar with Pascals triangle or binomial expansion?
Your answer is correct if it takes
8 yes votes 1 way
7 yes votes 8 ways
...

 

[pka]

Actually #3b is a bit more involved than the others
Gene is correct about “Pascal’s triangle or binomial expansion”
\(\displaystyle \L
\left( {\begin{array}{c}
N \\
k \\
\end{array}} \right) = \frac{{N!}}{{k!(N - k)!}}\).
To get five or more yes votes, select 5 or more places for Y in the other we can A or N. Therefore we have a total of \(\displaystyle \L
\sum\limits_{k = 5}^8 {\left( {\begin{array}{c}
8 \\
k \\
\end{array}} \right)2^{8 - k} }\) ways to have five or more Y's.

 

[desi43]

thank you so much you guys!