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wakoff
In how many ways can 8 employees be assigned to 3 identical offices? The answer is 1094, please explain how to get that answer.
Do all offices have to have at least one person in it?
i'm not quite sure, i suck at these word problems T__T
my teacher told me that it should look something like s(8,1) + s(8,2) + s(8,3)
Well, I figured it out both ways, and either way gives a lot more more than 1094 ways. Where did you get that number?
my teacher gave me that answer he also gave me another question Q: In how many ways can 8 employees be assigned to 5 identical offices? A: 3845
I'm not sure your teacher gave you the right answer on that first question, and I'l show you why so you can present him with what I believe is the right answer. So, we'll stick to the first question first. Here goes: ...
With any word problem, you can only make logical inferences. You can't read into the problem things that are not there. But you can make logical deductions. If things are unclear, then you have to make qualifications or LOGICAL assumptions. Let's assume for the moment that one or more offices can be unoccupied. Later, we'l get an answer for "at least one person in each office". It looks like these are the only 2 possiblities.
Here are the arrangements: 1) 8 0 0 2) 7 1 0 3) 6 2 0 4) 6 1 1 5) 5 3 0 6) 5 2 1 7) 4 4 0 8) 4 3 1 9) 4 2 2 10) 3 3 2 We have to talk about this for a second. First, notice that we don't have an arrangement 0 8 0 or 0 0 8, or for that matter 7 0 1 or 0 7 1 or other permutations, because we are taking "identical offices" at face value. So, it comes down to the various groupings. Study this list at great length when we are done because this will form the basis of your argument for your position.
Next, we can form 2 different answers with this listing, that being the first answer with "some unoccupied offices" and the second answer with "no unoccupied offices". So, 2 possible answers, but I would think that, if we take the question at face value, all ten arrangements are valid. If your teacher maintains "no unoccupied offices", then we leave out arrangements 1, 2, 3, 5, and 7, but we will still have more than 1094 for our answer.
So, we get the number of possibilities for each line. I'll write them down and then we can talk about them. Then just add them up! With or without leaving out what I described above.
The first employee has 3 choices of office, the second has 3 choices, etc. so the total combinations are: \[3^{8} = 6561\] This isn't even a multiple of your teacher's answer, so I have no idea how he got his answer.
i'm starting to think i copied the answer incorrectly
1) 8 0 0 1 2) 7 1 0 8 3) 6 2 0 28 4) 6 1 1 28 5) 5 3 0 56 6) 5 2 1 168 7) 4 4 0 70 8) 4 3 1 280 9) 4 2 2 420 10) 3 3 2 560 Now, we need to talk about each one separately. I know this looks like a LOT of work. and it is! But you will see a pattern emerge that will make this rather easy in concept, just a lot of work.
The first one is simple, there is only one way to get everyone in one room, because all 3 rooms are identical. This "identical" concept is key, otherwise we would have an astronomical number. So "identical" means something here. You follow?
i was thinking that it should be 8 employees where 1 is applied to each office for some reason
So, for #2, the arrangement 7 1 0, there are 8 ways to select the one guy "by himself" with only one way to get the rest of the folks into one room, so 8 x 1 or 8. Still with me?
It starts getting interesting now. for #3, the arrangement 6 2 0, there are 8C2 or 28 ways to get a pair with only one way to get the rest into one room. 8C2 is (8)(7)/(2)(1). Hope you're still with me, cuz this is fun now!
ohh... yes it's starting to make sense now
#4: 6 1 1 -> Very similar to case #3 except the "pair" is now in 2 rooms, but it is still logically a pair. Dwell on this for a moment. You'll have to see it and convince your teacher. It's a pair because there are 2 and it doesn't matter which one we select first, so its 8C2 again.
#5: 5 3 0 -> 8C3 because the "rest" are stuck in one room, so 56. I'm going to go a little faster now, so you'll have to stop me for fuller explanations, but I think you're getting this now. Go back and re-read everything whrn we're done and will argue it back and forth.
#6: 5 2 1 A little tricky. It's 8 times 7C2. This is 168. It comes from 8 from any of the 8 by himself TIMES any pair left from the 7. This is the trickiest so far, but it gets worse :-)
case #7: 4 4 0 8C4 which is 70 . This one was simple by comparison.
case #8: 4 3 1 -> 8 TIMES 7C3 = 280 Do you see the pattern and logic? Otherwise, stop me for explanation! 8 ways for the "by himself" times groups of 3 out of the remaining 7
Thank you so much i've been stressing over this problem all night (all nighter)
I actually have a question, why did you stop at specifically 10 arrangements?
case #9: more tricky! 4 2 2 -> 8C2 TIMES 6C2 = 420 It's just math, no sweat!
Before we get to #10, I'll stop for your question. To compile this table, we started , of course, with 8 0 0 and quickly went to arrangement(s) with 7 of which there are only one. Well, once you start getting to "3" in the first column, you start getting "repeats". For example, after doing 3 3 2, I could have written 3 2 3, but that was "done" already. Again, for example, if I went with 2 in the first column, all combinations are already taken if you rearrange the "2" selections from above. Do you follow that?
If I wrote 2 6 0, that's already case #3. If I wrote 2 5 1, that's already case # 6. And it is those cases, because of the "identical rooms". Again, taking the question at face value.
does that mean i can add up all of the computations and multiply that result by 3?
since they are identical?
that makes so much sense
So, our last: case #10: 3 3 2 -> 8C2 TIMES 6C3 = 560. Because after you pick 2 from 8, you have only 6 left from which to get a triplet. So, these are the completely differentiated , distinct cases which do comprise the different arrangements. Now we add them up. One of my posts near the top had all the numbers summarized so you can go from that post. And again, watch for the distinction of "no unoccupied rooms" or "unoccupied rooms allowable" and add up the numbers. BUT!!!!! even if you add up only 4, 6, 8, 9, and 10, you will get more than 1094. NOW! Let's discuss this, because you have to convince your teacher about this. AND, since I did so much work, I'd like you to message me back after you meet with your teacher. Agreed?
Even if you go with "no unoccupied rooms" and add up 4, 6, 8, 9, 10, you get 1456. I didn't add up the others, but if you include all TEN cases, the number is even bigger, and I think that the adding of ALL TEN cases gives the real answer. But tell your teacher about the two cases. And that he should listen to you.
i'm going to see him later today, expect a answer later tonight or tomorrow :)
You're welcome! And I DO make mistakes, more often than I like, but not here!
@tcarroll010 has the correct answer(s) only if the fact that the offices are identical means that 8 people in office 1 and no people in offices 2 and 3 is the same as 8 people in office 2 and no people in offices 1 and 3, etc. If that's not the case then my answer is correct.
I see now how the answer is 1094. I made a mistake in my number computations. I had written: original corrected correction made? 1) 8 0 0 1 1 2) 7 1 0 8 8 3) 6 2 0 28 28 4) 6 1 1 28 28 5) 5 3 0 56 56 6) 5 2 1 168 168 7) 4 4 0 70 35 Y 8) 4 3 1 280 280 9) 4 2 2 420 210 Y 10) 3 3 2 560 280 Y So, I needed to correct #'s 7, 9, and 10. The original values I had were double what they should have been and I should have seen that right off because each of those has a repeated number (4, 2, and then 3 for case 7, 9, and then 10). So, if you add up my corrected column, you will get 1094, just like the Stirling numbers type 2 give you. I never heard of the Stirling Numbers before and I'll have to read up on them. I did this the "long way", and you might want to go over my correction post here just to get an insight into how this has meaning. I could tell though, right off, that my method is definitely NOT the preferred method since it was so long and would be practical for only small numbers. This particular example is probably the largest anyone would ever want to do by hand or the "long way". So, what I did was really the "guts" of the Stirling number type 2 and it gives the right answer IF you make no mistakes!