#puzzle
A rearrangement of the letters of a word has no fixed letters if, when the rearrangement is placed directly below the word, no column has the same letter repeated. For instance, the blocks of letters below show that ESARET is a rearrangement with no fixed letters of T E R E S A, but R E A S T E is not.
T E R E S A T E R E S A
E S A R E T R E A S T E
How many distinguishable rearrangements with no fixed letters does T E R E S A have? (The two E’s are considered identical.)
Since we have been given a small search word, we could easily enumerate the combinations. For example, we can place the 2 E's in different position as follows:
E ? E ? ? ?
E ? ? ? E ?
E ? ? ? ? E
? ? E ? E ?
? ? E ? ? E
? ? ? ? E E
This yields 6 ways to do so. One way to think about this is that we are moving each of the E's to the remaining four positions and therefore there are 4-Choose-2 ways = 6 ways.
Taking one of the six arrangements above, we can also enumerate the possible arrangements of the remaining letters. Let us say E now occupies new positions X and Y. Neither X nor Y can be in their original position no matter where we put them. So we have to place the last two letters which we call M and N and there are only 4 positions to choose from. M can either be moved to where N was or it can be moved to one of the original positions of E. In the former case there are three positions remaining for N so there are 3 ways to do so. In the latter case there are two E's so there are two ways each for M and for N. There fore there are a total of 3 + 2 x 2 = 7 ways to place the last two letters. Moreover X and Y can also be interchanged so we have 7 times 2 equals 14 ways. Since we are doing this for each of the 6 ways to place E's to start with, we have a total of 6 x 14 = 84 ways. It is possible to enumerate all these 14 ways for a given starting arrangement of E and see that we can repeat the same for any of the remaining arrangements of E.
A rearrangement of the letters of a word has no fixed letters if, when the rearrangement is placed directly below the word, no column has the same letter repeated. For instance, the blocks of letters below show that ESARET is a rearrangement with no fixed letters of T E R E S A, but R E A S T E is not.
T E R E S A T E R E S A
E S A R E T R E A S T E
How many distinguishable rearrangements with no fixed letters does T E R E S A have? (The two E’s are considered identical.)
Since we have been given a small search word, we could easily enumerate the combinations. For example, we can place the 2 E's in different position as follows:
E ? E ? ? ?
E ? ? ? E ?
E ? ? ? ? E
? ? E ? E ?
? ? E ? ? E
? ? ? ? E E
This yields 6 ways to do so. One way to think about this is that we are moving each of the E's to the remaining four positions and therefore there are 4-Choose-2 ways = 6 ways.
Taking one of the six arrangements above, we can also enumerate the possible arrangements of the remaining letters. Let us say E now occupies new positions X and Y. Neither X nor Y can be in their original position no matter where we put them. So we have to place the last two letters which we call M and N and there are only 4 positions to choose from. M can either be moved to where N was or it can be moved to one of the original positions of E. In the former case there are three positions remaining for N so there are 3 ways to do so. In the latter case there are two E's so there are two ways each for M and for N. There fore there are a total of 3 + 2 x 2 = 7 ways to place the last two letters. Moreover X and Y can also be interchanged so we have 7 times 2 equals 14 ways. Since we are doing this for each of the 6 ways to place E's to start with, we have a total of 6 x 14 = 84 ways. It is possible to enumerate all these 14 ways for a given starting arrangement of E and see that we can repeat the same for any of the remaining arrangements of E.
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