Basic Stats Question! Plz help

[somebody]

hey guys, i need help with stat question. This is one of those extra questions on the back of the book that is possibly going to help me in my understanding:

 

So question is:

 

There are 5 boys and 5 girls in a group that forms a line. In how many

ways can you arrange them in a line if at one end of the line there must be

a boy and at the other end there must be a girl?

 

I was thinking it would be something like: at one end you have 5 possibilities for a boy to be there and at the other end there is 5 possibilities for a girl to be there. Since there are total of 10 people and 2 of them will be at the either end, then we have 8 people left. So in the middle it would be like 8 factorial. So overall, it would be something like 5*8!*5?

 

A buddy of mine suggested that it would be 8!*2!. He made his arguement by saying its a restrictive permutation. I googled it but nothing came up so i came here.

 

Please help and do explain your answer if its 8!*2!.

 

Thank You,

Quick response would be great!!!

[Jay-qu]

Well it all depends on whether the 5 girls are distinguishable or not (same goes with the guys obviously). If they are indistinguishable then there is only 2 possibilities for 'what' can be standing in each place - and the ends become irrelevant because they are set.

[Essay]

I think you'd have to assume they are indistinguishable.

===

 

I got an A in Stat 400, but that was 30 years ago (and one of the few A's)....

So I don't think you should quote this....

I'd like to see some real math folks tackle this one first.

===

 

...but imho....

Wouldn't factorial be used if you had 10 different genders (distinguishable boys and girls)?

 

Isn't this just a case of how many way can you arrange 2 different kinds of objects in an array of 8 places

...so it'd be 2 raised to the number of different permutations--(or words to that effect).

 

Really you're just dealing with the 4 boys and 4 girls in the interior of the line, right?

So this would be 2 to the 8th power, plus 2--for the two different end-position options.

(2^8) + 2 = ?

 

ummmm...

...or would it be (2^4) + 2 = ?

 

Thankfully, that was 30 years ago--when I didn't care if my head hurt a little bit!

[Turtle]

hey guys, i need help with stat question. This is one of those extra questions on the back of the book that is possibly going to help me in my understanding:

 

So question is:

 

There are 5 boys and 5 girls in a group that forms a line. In how many

ways can you arrange them in a line if at one end of the line there must be

a boy and at the other end there must be a girl?

 

I was thinking it would be something like: at one end you have 5 possibilities for a boy to be there and at the other end there is 5 possibilities for a girl to be there. Since there are total of 10 people and 2 of them will be at the either end, then we have 8 people left. So in the middle it would be like 8 factorial. So overall, it would be something like 5*8!*5?

 

A buddy of mine suggested that it would be 8!*2!. He made his arguement by saying its a restrictive permutation. I googled it but nothing came up so i came here.

 

Please help and do explain your answer if its 8!*2!.

 

Thank You,

Quick response would be great!!!

 

I think your 5*8!*5 is right, at least as long as the line does not have to be formed starting at one end and choosing 1-by-1 until reaching the end.

 

Then me thinks it would be something like 5 ways to make the first choice, 9 ways to make the second, 8 the 3rd, 7 the 4th, ...and then I start loosing it because you have to start considering if you have any of the opposite you started with to go at the end. :naughty:

 

So much for quick, and as Essay says, we can wait for the real math guys to post because we can always count on them to keep all our ducks in a row. ;)

[freeztar]

I'm with you Essay on the memory issue...

 

I'll venture a guess and say:

 

[math]10!/2[/math]

 

Good question!

[Essay]

But the first thing to figure out is whether or not it just all the possible boy/girl arrangements, or if it is all arrangements of 10 different people (just avoiding the boy/boy or girl/girl "bookends").

[Jay-qu]

If they are indistinguishable, then it is just like binary counting. Like Essay said - it should be 2^8+2

[somebody]

Hey guys, I had my midterm and that question did show up. Fortunately, professor messed up on the wording so many kids got confused so i think he might just wave that question and give everyone full credit.

 

But to clarify things, those boys and girls are indistinguishable from each other. I am happy that this is making people think way back from the college courses. But any of you interested in further evaluating this problem then here is a great helper:

 

Permutation and combination tutorial: Restricted Permutations

[Jay-qu]

Hey guys, I had my midterm and that question did show up. Fortunately, professor messed up on the wording so many kids got confused so i think he might just wave that question and give everyone full credit.

 

But to clarify things, those boys and girls are indistinguishable from each other. I am happy that this is making people think way back from the college courses. But any of you interested in further evaluating this problem then here is a great helper:

 

Permutation and combination tutorial: Restricted Permutations

Yup way back (2 years ago :))

[CraigD]

I don’t see that this thread quite reached a satisfying conclusion, so will add my bit to it.

There are 5 boys and 5 girls in a group that forms a line. In how many

ways can you arrange them in a line if at one end of the line there must be

a boy and at the other end there must be a girl?

 

I was thinking it would be something like: at one end you have 5 possibilities for a boy to be there and at the other end there is 5 possibilities for a girl to be there. Since there are total of 10 people and 2 of them will be at the either end, then we have 8 people left. So in the middle it would be like 8 factorial. So overall, it would be something like 5*8!*5?

This answer, [math]5 \cdot 5 \cdot 8! = 1008000[/math], is correct for the case where each boy or girl is distinct, and the line either always starts with a boy, or always starts with a girl. The question appears to me to state that one can chose either a boy or girl for the start of the line, in which case there are 10, rather than 5 choices for the first position, doubling the to number of possible arrangements to [math]10 \cdot 5 \cdot 8! = 2016000[/math].

 

If the boys are indistinguishable from one another and the girls are indistinguishable from one another, it’s necessary to divide the number by the number of possible arangements with each indistinguishable collection, getting [math]\frac{10 \cdot 5 \cdot 8! }{5! \cdot 5!} = 140[/math].

[maddog]

hey guys, i need help with stat question. This is one of those extra questions on the back of the book that is possibly going to help me in my understanding:

 

So question is:

 

There are 5 boys and 5 girls in a group that forms a line. In how many

ways can you arrange them in a line if at one end of the line there must be

a boy and at the other end there must be a girl?

 

I was thinking it would be something like: at one end you have 5 possibilities for a boy to be there and at the other end there is 5 possibilities for a girl to be there. Since there are total of 10 people and 2 of them will be at the either end, then we have 8 people left. So in the middle it would be like 8 factorial. So overall, it would be something like 5*8!*5?

 

A buddy of mine suggested that it would be 8!*2!. He made his arguement by saying its a restrictive permutation. I googled it but nothing came up so i came here.

 

Please help and do explain your answer if its 8!*2!.

I would go with your and Turtle's value of (5^2)8!. It is as your friend said a restricted permutation. I disagree with his answer though.

 

Line up ten boxes. In each end box has only 5 choices (twice) = 5^2. the middle boxes are eight boxes without restrictions on their arrangement (8!) possibilities.

 

I hope this helps you to understand more.

 

:weather_snowing:

 

maddog

[CraigD]

I would go with your and Turtle's value of (5^2)8!. It is as your friend said a restricted permutation. I disagree with his answer though.

 

Line up ten boxes. In each end box has only 5 choices (twice) = 5^2. the middle boxes are eight boxes without restrictions on their arrangement (8!) possibilities.

I think you’ve fallen for the question’s “trick”, if it can be said to have one, understating the number of possible arrangements by a factor of 2.

 

Your choice for the first box is not from 5, but from 10, because you may chose any of the 10 people, regardless of gender. Your next choice, for the last box, is from the 5 people of the gender you didn’t pick in your first choice. So the answer is 10 * 5 * 8! = 2016000.

[maddog]

I think you’ve fallen for the question’s “trick”, if it can be said to have one, understating the number of possible arrangements by a factor of 2.

 

Your choice for the first box is not from 5, but from 10, because you may chose any of the 10 people, regardless of gender. Your next choice, for the last box, is from the 5 people of the gender you didn’t pick in your first choice. So the answer is 10 * 5 * 8! = 2016000.

I often misunderstand such subtleties. My assumption was boy was on a specific end as was the girl. If either end suffices for either gender then I agree with your analysis.

 

maddog