Probability - please help!

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Kiwigirl

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Feb 27, 2006
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Can anyone please help me answer this?

1. Six different letters are to be put into 6 addressed envelopes.
a) In how many ways can this be done?
b) In how many ways can letters be put into envelopes with the wrong address?

2. In how many ways can 10 chairs be allocated to three rooms with 3 in the first, 5 in the second, and 2 in the third?
 
You can put any of 6 in the first.
You can put any of the other 5 in the second.
You can put any of the other 4 in the third.
Etc.
6*5*4*3*2*1 = 6! (Six factorial)
There is only one right way so there are 6!-1 wrong ways.
 
Here are two questions.
In 1b, do you mean that each letter gets into to wrong envelope?
Or does it mean at least one is in the wrong envelope? This is Gene’s answer!

In 2, are we to assume that the chairs are all different?
If they are not different, then the answer is 1.
 
Answer to pka

Thank you for your help.
1a still has had no response and I'm not sure of the answer to this question. In 1b, the letters are addressed, so the question asks in how many ways the letters can be put into the wrong envelopes (not the correct envelope for the addressed letters). In 2, the question doesn't say that the chairs are different, so I assume that they are the same but that there are three different rooms for them to be placed into. It asks, in how many ways can 10 chairs be allocated to three rooms with three in the first, five in the second and two in the third room.
 
Gene said:
You can put any of 6 in the first.
You can put any of the other 5 in the second.
You can put any of the other 4 in the third.
Etc.
6*5*4*3*2*1 = 6! (Six factorial)
There is only one right way so there are 6!-1 wrong ways.
Click to expand...

Hi Gene, thanks for your quick response. I'm still not sure how to answer the other questions though, do you have any idea? Also, pka says that you have answered the wrong question, with one letter going into the wrong envelope? Sorry, I know I must seem dumb, but probability has never been my good subject. Thank you for any further help :)
 
Kiwigirl said:
pka says that you have answered the wrong question, with one letter going into the wrong envelope?
Click to expand...
NO! I did not say that.
I said that if question means that at least one is in the wrong envelope then the answer Gene gave is correct. But if the questions means that all the letters endup in the wrong envelope then that is not the answer.
 
pka said:
Kiwigirl said:
pka says that you have answered the wrong question, with one letter going into the wrong envelope?
Click to expand...
NO! I did not say that.
I said that if question means that at least one is in the wrong envelope then the answer Gene gave is correct. But if the questions means that all the letters endup in the wrong envelope then that is not the answer.
Click to expand...

Yes, I understand that pka. It is true that all of the letters end up in the wrong envelope, this is why I said that it is wrong.
 
1b) is a head-scratcher.
If there were 2 envelopes there would be 1 way.
If there were 3 envelopes there would be 2 ways.
If there were 4 envelopes there would be 9 ways. I counted those to see what is happening.
I'm sure the answer has a couple factorials and a couple linear terms but I can't come up with an equation that works even on 3 envelopes.
My computer says that if there were 5 envelopes there would be 44 ways and 6 gives 309 ways but I doubt that your teacher would accept a copy of my simple-minded program. It gave correct answers for 2,3 & 4 so I hope it is also correct for 5 & 6.

2) As PKA said IF the chairs are interchangable there is only one way. If they are numbered so chair 1 in room 1 is different than chair 1 in room 2 I think the formula is 10!/(2!*3!*5!)
 
The answer to 1b) is a bit advanced. These are call derangements: the total rearrangement in which no term is in its proper place.
The string “123456” rearranged as “645231” is a derangement. But “653241” is not because of the ‘3’.
The number of derangement of a N-string is \(\displaystyle \L
D(N) = (N!)\sum\limits_{k = 0}^N {\frac{{\left( { - 1} \right)^k }}{{k!}}}\)
\(\displaystyle \L
D(6) = 265\) that is the answer to 1b. None of the letters is in the correct envelops.

Gene is correct about #2.
 
Oooops, for 6 envelopes I miss-copied my scratched down result. 265 is correct.

PS. My program counted 654321. Typo?
 
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