<p>what is the fastest way to solve this problem: Ann, Benito, Chantra, and Desiree are each to be assigned to 1 of 2 vehicles, a bus or a van. How many different assignments of these 4 people are possible if at least one person must be assigned to each vehicle?</p>
<p>here’s how I did it:</p>
<p>I first found the total number assignments for these four students. There are 4<em>3</em>2*1=24 possible assignenments (you have four people to choose from at first, then three, then two, then one, placing them wherever you want). Right now we are not concerned with where these people end up. </p>
<p>Then, I considered the probability. To make things easier, I wanted to find the probability that all four people end up in the same vehicle (which is the same as the negation or 1-the probability that there is at least one person in each vehicle). There is a (1/2)<em>(1/2)</em>(1/2)*(1/2)=(1/2)^4=1/16 probability of all four being assigned to the van (think of it as flipping coins to determine where they go). There is also a 1/16 probability that all four are assigned to the bus. Hence, there is a 1/16+1/16=2/16=1/8 probability that all four are in the same vehicle, thus a 7/8 probability that at least one is in each vehicle. </p>
<p>Now, remember that there were 24 possible ways to assign. Only 7/8 of these will result in at least one student being in each vehicle, so there are 24*(7/8) or 21 possible assignments. </p>
<p>Right? There’s probably a quicker way (maybe something with the choose function). I was trying to come up with some faster solutions but couldn’t. This is a harder problem then I’m used to seeing. Very conceptual, imo.</p>
<p>[edit] LEt me look at this again.</p>
<p>by the way the right answer is 14</p>
<p>Uh, 4ncr3<em>1ncr1 + 4ncr3</em>1ncr1+ 4ncr2*2ncr2 =14</p>
<p>4ncr 3 is the possibilities of putting 4 people into 3 spots, the last one in another. Goes both ways for the bus and van.</p>
<p>4 ncr 2 = is the possibilities of putting 4 people into 2 spots. the other 2 go into the other vehicle.</p>
<p>^OK, I’m pretty convinced 14 is right. Not sure where my strategy went wrong. I feel kinda dumb, actually. That was pretty simple. I think I went wrong by thinking of trying to find the assignments for all 4 people in the same vehicle (which you can’t really do choose for for multiple reasons).</p>
<p>I was trying random stuff and it didn’t work. So I found one that matches his answer and made a reasonable explanation and it makes sense. :D</p>
<p>I came to the conclusion for both 4ncr3 + 4 ncr3, but I wasn’t sure about the 4ncr2, but it makes sense now that I think about it.</p>
<p>Okay, actually the 4ncr2 part is a bit questionable. It makes sense, but it’s a bit weird.</p>
<p>What was wrong with my approach, though?</p>
<p>Using the FCP:</p>
<p>For a van you have 4 choices. For the car you have 3 choices left. This is 12.
Now it doesn’t matter what spots the last two people take. You have 2 choices for the first and 1 choice for the last.</p>
<p>(4<em>3) + (2</em>1)=14</p>
<p>I’m not sure, I don’t like to think about probability subjectively. Just add up the possibilties and think as little as possible. O_o</p>
<p>The above method works, I don’t know why because 4npr3 = 12, which is 4*3, but this is definitely combinations.</p>
<p>@Grozza</p>
<p>Your approach fails because there is not 24 possible ways to assign…</p>
<p>GRE why does your method work?</p>
<p>4ncr3 + 4ncr3+4ncr2, makes complete sense using combinations and stuff…You’re using permutations… I don’t know how that works…</p>
<p>You have to meet the constraint of at least 1 person being assigned to each vehicle.</p>
<p>You have 4 people for the 1st vehicle and 3 people for the 2nd, by the FCP there are 12 possibilities. </p>
<p>Now you have 2 people left and it doesn’t matter which vehicle they are assigned to. There are 2 ways (2*1) to assign them to those vehicles.</p>
<p>So you have 14 possibilities.</p>
<p>(BTW cjgone I figured out the milk/cream problem, I posted it in my thread.)</p>
<p>thanks for the help.
i think i figured out another way to do it as well: forget about the constraint temporarily and do the permutations of 4 people with the possibility of having 4 people in one van; you get 4 x 4 which is 16. then, you apply the constraint of at least 1 person per vehicle and subtract the two possible situations in which there would be 4 people in one vehicle. So you get 16 -2 which is 14.</p>
<p>^There we go! My way was somewhat valid. I guess I just found out the total assignments wrong. So, if the total assignments is 16, then the probability method I did would still hold and one would get 14 (but you’re way would be quicker at that point). I was wondering how to get the 16 total assignments and I guess you figured it out (I still don’t get how you get 16, though. my combinatorics is rusty, to say the least. can you explain that and why you’d permute this case? I’m kinda interested in clearing this up in my head).</p>
<p>disregarding the rule, you have two spots and 4 people for each spot because one of the vehicles can be empty, then you subtract the two cases where there are 4 people on one vehicle.</p>
<p>^yeah but how’d you get 16? 4p2? 4p2 + 3p2? 2*4p2? etc?</p>
<p>just 4 times i dont use the cr method i just use boxes and multiply</p>
<p>just 4 times 4: i dont use the cr method i just use boxes and multiply</p>
<p>Here’s how I did it</p>
<p>4 possible ways when one person is in the bus
4 possible ways when one person is in the van
4 nCr 2 = 6 possible ways of putting 2 people in the bus (the other two “fall” into place)</p>
<p>4 + 4 + 6
= 14</p>