IMP Year 2 POW 1

M

Magnus

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Sep 26, 2005
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Ok, I'm new here, and was wondering if anyone could tell me if I was correct in finding that there is only one combination for the 5 places, and the only one that I found was

21200

If you are trying to figure out, what I am talking about, I will elaborate-

Ok there are 5 places like so (down below) for you to place a number of how many times you are going to use that number, I need to find all the possible combinations there are for the five places, so if you place a two in the zero place, you plan on using zero twice


__ __ __ __ __
0 1 2 3 4

So I got

2 1 2 0 0
0 1 2 3 4

Am I correct, please let me know, and if you would be kind enough to explain how I would be able to tell how I know that this is or is not the total amount of combinations, and if would be willing to share any combonations you have found, I would really appreciate it, by the way heres a rule that I have found, the total amount of boxes, is what the sum of the numbers should be, so 2+1+2 is five and there are five boxes.
 
You forgot to include the statement of the question. For what is "21200" an answer?

Eliz.
 
Magnus, Good grief, that is a confusing posting.
Are you asking: “How many ways can one use five of the digits 0,1,2,3,4 to add up to 5.?”
Could one way be 1+1+1+1+1?
Or 0+0+1+1+3?
Yes what is the question?
 
No, the question how many possible combonations are there using those numbers, look at the first on I put up> I'll explain the 21200

Ok, you have the 5 spaces listed from 0-4. So I placed a "2" in the zero place stating that I am going to use it (Zero) twice. I placed a "1" in the one's place stating that I am going to use it once, I placed a "2" in the twos place stating that I am going to use it twice, and so on with a zero in the three's and fours' place that I am never going to use it, you can put any of those five numbers in any place, just make sure that it works, and that you can use that number as many times you have listed for it, so if you place three in the fours place, you have to use 4 three times, and then you have to place a 1 in the threes place because you used that once so far, you have to kep going until its balanced out. do you understand now?
 
Magnus said:
No, the question how many possible combonations are there using those numbers....do you understand now?
Click to expand...
:shock:

No. Please reply with the full and exact text of the exercise, just as it was provided to you. Thank you.

Eliz.
 
Why is it that people have such reluctance to posting the actual problem?
Giving us what you think it means is not very helpful. For example:
“you can put any of those five numbers in any place”: WHAT FIVE NUMBERS?
“just make sure that it works”: What does it mean to work?
“and then you have to place a 1 in the threes”: WHAT?
“keep going until its balanced out”: balanced what?
Please just give us the actual statement of the question. PLEASE
 
The problem was clear to me, though a neat answer isn't.
Write a 5 digit number using 0 to 4 such that
the first digit is how many zeros are in it,
the second digit is how many ones are in it,
the third digit is how many twos are in it,
the forth digit is how many threes are in it,
the fifth digit is how many fours are in it.
I've only gotten a chain of logic that eliminates 0,1 and 4 as the first digit like:
It can't be zero 'cause it has at least one zero when you say there are zero zeros.
It can't be 1 'cause the second can't be 0 or 1 without a contradiction and can't be 3 or 4 'cause... etc.
I don't see any other way but it becomes a very branchy tree. I think he's right. 21200 is the only valid number, but the proof is too long.
 
Those are known as Self-Descriptive Numbers.
A 10digit one:

0123456789
6210001000

6 zeroes
2 ones
1 two
0 threes
0 fours
0 fives
1 six
0 sevens
0 eights
0 nines
 
Denis, thank you very much.
“Those are known as Self-Descriptive Numbers.”
That makes perfect sense.
I admire Gene for being able to decipher the problem.
I admit to never having seen this idea.
 
Still thinking. It's not as hard as anticipated if my logic is good.
First digit can't be 4 'cause it would have a 4, four zeros and a 1 for the 4. Wouldn't fit.
First digit can't be 3 'cause it would have a 3, three zeros and a 1 for the 3 and another for the 1. Wouldn't fit.
First digit IS 2.
That leaves 0,1 or 2 for the second.
It can't be 2 'cause it has two 0s,two 1s and the two 2s from 22???. Wouldn't fit.
If it's 1, that forces your answer.
The only thing left is 20???. The third digit must be 2 or 3 'cause of no 1s and one 2.
I'm sure you can take it from there.

BTW I had never seen it this way but
In this sentence there are ___ zeros, ___ ones, ___ twos, ___ threes ___ fours... ___ eights and ___ nines.
Click to expand...
Fill in the ___s with integers so the sentence is true.

I've never tried it as
In this sentence there are ___ 0s, ___ 1s, ___ 2s, ...
 
While we try yours Gene, try this one of mine:

After entering the number of times that each vowel appears in
this sentence, you should end with A a's, E e's, I i's,
O o's and U u's, else you goofed so start over.

Note: "entering" the number means the written out number, like "two a's".

solution down South; DON'T LOOK!
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After entering the number of times that each vowel appears in
this sentence, you should end up with eight a's, twenty four e's,
nine i's, eleven o's and six u's, else you goofed so start over.
 
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