You are not logged in.
Pages: 1
There are 8 students in a class, including Gina and Christina. In how many ways can you arrange them in a row so that Gina and Christina are always next to each other? Show your work in detail.
My thinking, is that there are 8 places for Gina to be seated, and then 2 spots for Christina to be seated. This leaves 6! places for the rest of the people to be seated. Is that correct?
Offline
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
So you're saying that to solve this problem, first pretend that Gina and Christina are the same people, and then find a way to arrange 7! people?
Offline
I think that overcounts. You were close with your first answer but overcounted there too. Assuming that the six remaining students are distinguishable I am getting 14 * 6! = 10080
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
No overcounting. 7!*2! ; 2! for permutation within the rope.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Thickhead is right. Let's first seat all the "noname-students" randomly, there are 6! ways to do that. Now we choose a random spot for our Frankenstein's Monster (aka both of them tied together), which can be in one of 7 places, i.e. 7C1 (which is just 7), so we get 6!*7 = 7!. Now we can double that count for a) Gina-Christina and b) Christina-Gina.
So we get 7!*2, which is obviously the same as 7!*2!.
Edit: Just to clarify why the opening post is wrong: You assume there are 8 different spots for Gina to be seated, but you can't sit Gina to the right (or left) of Christina, so there are actually only 7.
Last edited by Lehona (2016-10-23 00:44:34)
Offline
In how many ways can you arrange them in a row
In arrangements order counts. _ _ c g _ _ _ _ and _ _ g c _ _ _ _ are different.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
That's what the factor 2 is accounting for (cases a) and b) in my post).
Offline
Hi;
I was just adding something to my method for evene. See my post #4. You have the same answer I do.
I do not thickhead noticed that.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Permutations always go by logic. Whatever be the logic finally they should converge.But the logic must be clear and not be too long winded. If 3 people are to sit together also I use the same logic 6!*3!
Last edited by thickhead (2016-10-23 03:39:09)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Pages: 1