MHB How many possible ways can this be done? (A,B,C,D,E)

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The discussion centers on the arrangement of letters A, B, C, D, and E, with the goal of determining the number of possible combinations that include at least one letter. It concludes that there are 31 valid combinations, derived from the formula $2^5 - 1$, which accounts for all possible configurations minus the empty set. Participants inquire about the existence of online calculators for such problems and seek a specific name for this type of combinatorial problem. The method used involves counting the arrangements in a structured format for clarity. Overall, the focus is on understanding the combinatorial possibilities of the given letters.
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Example:
1. (A,B,C,D,E)
2. (A,B,C,D)
3. (A,B,C)
4. (A,B)
5. (A)
6. (A,B,D,E)
etc...
I just put it in a table format, as it's easier to count this way (31 columns = 31 solutions).
Not sure if there are any more ways to do this.

Questions:

1) Are there only 31 ways to do this?

2) Is there like a calculator online which solves this sort of problem?

3) Is there a name for this sort of problem?
 
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It appears you want A, B, C, D and E arranged in alphabetical order with at least one letter present. There are $2^5-1 = 31$ ways to do this; we subtract $1$ to omit the configuration of having all blanks. The $2^5$ comes from the number of possibilities for each "slot", which is $2$: a letter or a blank.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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