Coin PatternsDate: 08/26/2003 at 23:22:10 From: Dawn Subject: Patterns There are four rows of coins. Each row has exactly one penny, one nickel, one dime, and one quarter. No row, either horizontal, vertical, or diagonal, has more than one coin of each kind. How are the coins arranged? I know there is a formula but I can't remember it. Date: 08/28/2003 at 10:42:47 From: Doctor Ian Subject: Re: Patterns Hi Dawn, There isn't a formula, actually. But there are systematic ways you can approach this kind of problem. I'm going to use a, b, c, and d to represent the coins, just because they're easier to think about than P, N, D, and Q. At first, any coin could be anywhere: abcd | abcd | abcd | abcd -----+------+------+----- abcd | abcd | abcd | abcd -----+------+------+----- abcd | abcd | abcd | abcd -----+------+------+----- abcd | abcd | abcd | abcd For fun, let's make the first row a-b-c-d. Since we haven't really decided which coins are represented by which letters, there's no way for this to be wrong. a | b | c | d -----+------+------+----- abcd | abcd | abcd | abcd -----+------+------+----- abcd | abcd | abcd | abcd -----+------+------+----- abcd | abcd | abcd | abcd Having made this choice, we can rule out some other possibilities. For example, we can't have an a in the first place in any other row, right? Similarly, we can't have b, c, or d in the second, third, or fourth places: a | b | c | d -----+------+------+----- bcd | a cd | ab d | abc -----+------+------+----- bcd | a cd | ab d | abc -----+------+------+----- bcd | a cd | ab d | abc We also know that a and d can't appear elsewhere in the diagonals: a | b | c | d -----+------+------+----- bcd | cd | ab | abc -----+------+------+----- bcd | a c | b d | abc -----+------+------+----- bc | a cd | ab d | bc Now we've narrowed things down some, but not enough to get a unique solution yet. Let's look at the final row. How many possibilities remain? The first place can be b or c: b _ _ _ c _ _ _ The second place can be a, c, or d; except we can't repeat a letter within the row: b a _ _ b c _ _ b d _ _ c a _ _ c d _ _ The third place can be a, b, or d, again without repeats: b a d _ b c a _ b c d _ b d a _ c a b _ c a d _ c d a _ c d b _ In each case, there's only one possible letter left, if we can't have repeats: b a d c b c a d b c d a b d a c c a b d c a d b c d a b c d b a But the final letter has to be a b or a c, so we can throw away any row where this isn't true: b a d c b d a c c a d b c d a b That leaves us with only four possibilities. Now, maybe only one of these will work. Maybe all four will work. The only way to find out is to try each one of them. For example, let's try 'badc' in the final row: a | b | c | d -----+------+------+----- bcd | cd | ab | abc -----+------+------+----- bcd | a c | b d | abc -----+------+------+----- b | a | d | c Now we can rule out those same letters in the same places in the other rows: a | b | c | d -----+------+------+----- cd | cd | ab | ab -----+------+------+----- cd | c | b | ab -----+------+------+----- b | a | d | c Now we know where c and b have to be in the third row, so we can remove them everywhere that would cause a conflict (i.e., elsewhere in the same row and column): a | b | c | d -----+------+------+----- cd | d | a | ab -----+------+------+----- d | c | b | a -----+------+------+----- b | a | d | c And now we know where a and d must be in the second row, so we can do the same thing: a | b | c | d -----+------+------+----- c | d | a | b -----+------+------+----- d | c | b | a -----+------+------+----- b | a | d | c Now, the question is: Do the other possibilities also work, or is this the only one? Can you work that out for yourself? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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