Question 1137632: Sit 5 boys and 7 girls in a row. What is the probability that either all boys or all the girls sitting together?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Scenario 1: All boys sit together. The girls can sit however they want (either all girls are together or they are not all together).
In this scenario, we can think of 8 blocks labeled A, B, C, D, E, F, G, H
Each block is either a single seat, or it is a collection of five seats combined. What I've done is group all the "5 boys" into one single "seat", so to speak, and giving it a label. Let's say that this label is "label A". The other 7 labels B through H represent a single seat for each of the seven girls.
One permutation would be {A, B, C, D, E, F, G, H} which is probably naturally first to think about, as it is in alphabetical order. Another permutation would be {A, G, C, D, E, F, B, H}. I swapped letters B and G for the second permutation. The question is: how many total permutations are there? To answer this question, we consider how many letters we have for the first slot. That would be 8 choices total. Then after we make a selection, we have 8-1 = 7 left over for the second slot. Then 8-2 = 6 for the third slot. And so on.
Overall there are 8*7*6*5*4*3*2*1 = 40320 different ways to arrange those 8 letters. Order matters.
Through shorthand notation, we write 8! = 40320. The exclamation mark is commonly used for factorials. A factorial is where you start with some whole number and count down to 1 while multiplying your way through the countdown.
Recall that I made "block A" to represent the 5 boys. There are 5! = 5*4*3*2*1 = 120 different ways to arrange just the boys within any single block. So we multiply 120 by the previous result 40320 to get 120*40320 = 4,838,400
So overall there are 4,838,400 different ways to arrange all 12 students such that the boys must stick together in a single group. The girls are free to choose whether to be on their own or be with the group of other girls (or form a smaller sub group away from the main group of girls).
We will use this value later, so let's make M = 4,838,400
Now consider the restriction that "the boys must sit together" to be lifted. Now any student has total freedom to choose where they wish to sit, of course keeping in mind that they can't choose a seat already taken. How many ways are there to permute the 5+7 = 12 students? That would be 12! = 12*11*10*9*8*7*6*5*4*3*2*1 = 479,001,600
There are 479,001,600 different ways to arrange all the students where order matters, but boys don't have to sit together. We'll use this value later so let's make N = 479,001,600
At this point, we divide the values of M and N
M/N = (4,838,400)/(479,001,600) = 0.0101 approximately
It turns out that "01" decimal sequence goes on forever, so it allows us to convert that result to the fraction 1/99
So the probability of all boys sitting together is 1/99 = 0.0101
----------------------------------------------------------------------------------------------------
Scenario 2: All girls sit together. The boys are free to choose whether they all sit together or not.
We now have 6 blocks
A,B,C,D,E,F
Block A is the block of 7 girls combined as one "seat", while B through F are the five individual seats for the five boys.
There are 6! = 6*5*4*3*2*1 = 720 ways to arrange those 6 blocks and 7! = 7*6*5*4*3*2*1 = 5040 ways to arrange just the girls in any single block. So there are 720*5040 = 3,628,800 ways to arrange the students so that the girls must sit together (the boys do not). Let's call this Q = 3,628,800 so we can use it later.
This is out of N = 479,001,600 different ways to permute the 12 students without any seat restrictions. We computed this back in scenario 1.
So we get
Q/N = (3,628,800)/(479,001,600) = 0.00757575757576
which converts to the fraction 1/132
So the probability of all girls sitting together is 1/132 = 0.00757575757576
----------------------------------------------------------------------------------------------------
Scenario 3: The boys sit together. The girls sit together. No boy or girl can sit on their own, or can form a smaller sub-group.
Scenario 3 is only possible if we have two blocks: A, B
There are 2 ways to order these blocks.
If block A represents the boys, then we have 5! = 120 ways to arrange the boys in this block.
There are 7! = 5040 ways to arrange the girls in block B
Overall, there are 2*120*5040 = 1,209,600 different ways to arrange the students so the two genders stay in their respective groups. The 2 out front is to reflect that there are two permutations in {A,B}. So the boy block could go first and we'd have AB, or we can have the girl block seated first to have BA instead.
Let's make R = 1,209,600
Divide this over the value of N found earlier to get
R/N = (1,209,600)/(479,001,600) = 0.00252525252526
This converts to the fraction 1/396
----------------------------------------------------------------------------------------------------
That's a lot of work for those three scenarios, but hopefully you made it this far. The results we got were
- Scenario 1: Probability of all boys sitting together (girls don't have to be grouped) = 1/99
- Scenario 2: Probability of all girls sitting together (boys don't have to be grouped) = 1/132
- Scenario 3: Probability of both genders fully separated = 1/396
We'll add the probabilities of scenarios 1 and 2. Then subtract off the probability of scenario 3, as this scenario effectively double counts parts of the previous two scenarios. We get the following:
1/99 + 1/132 - 1/396 = 4/396 + 3/396 - 1/396
1/99 + 1/132 - 1/396 = (4+3-1)/396
1/99 + 1/132 - 1/396 = 6/396
1/99 + 1/132 - 1/396 = 1/66
The final answer as a fraction is 1/66
This approximates to the decimal value 0.01515 where the "15" sequence goes on forever
So this is roughly a 1.515 % chance to have the boys sit together, or the girls sit together, or both genders are separated in their respective blocks.
|
|
|
