Questions 1
Answers 32


After careful consideration of this question I now knoweth this:

Her Majesty’s solution to this be correct.

Sir CPhill’s subsequent consideration (of duplicates) to the solution be also correct.

There be two finite samples of symbol of differing size. One be size (26) and one be size six (6). Further, there be 7 spaces for them to occupy, and four spaces must be a letter symbol and three spaces must be a number symbol.

To reiterate: There must be 4 letters (A-Z) and there must be 3 numbers (0-5). Their location in the sequence be unknown. The letters and numbers be repeatable.

Here be why. Consider that there must be four letter symbols (of 26) and 3 number symbols (of 6), because this always be a requirement the space for them to occupy always remains the same. But the tricky part be the order, which Her Majesty discerneth from the beginning. * All these be possible, and yieldeth the same value:(26*26*26*26*6*6*6) or (6*6*6*26*26*26*26) or (26*6*26*6*26*6*26) or (6*26*26*6*26*26*6) and these all factor to (26^4) and (6^3) and giveth 98,706,816. (This solution would only be true if the positions were always known, no matter what the positions be).

HOWEVER the order matters, for the output be different. Consider this arrangement: (L1,L2,N1,N2,L3,L4,N3) Here the first 2 will only be letters (A-Z), the second two will only be numbers (0-5), and Ls will only be letters and Ns will only be numbers. If the sequence be (N1,L1,N2,L2,L3,N3,L4) a different output occureth though the total potential quantities remain the same.

To solve this, the permutation be used. There be 7 symbols and its factorial be 7! (5040) There be 5040 different ways to arrange these symbols . However, the 4 L’s be indiscernible from each other and the 3 Ns be indiscernible from each other, for they can generate the same values. The output would be the same if L1 be switched with L3 or any other L; so too, the Ns be transposable, without consequence.

HOWEVER, It would matter if an L switched with an N. To solve this, 7! be divided by the factorial of each indiscernible, these be the “4 Ls” and “3 Ns”. 7!/(4!*3!)=35. This giveth a discernible permutation and this (35) be multiplied by (26^4)* (6^3) to yield all discernible permutations.This be the more formal statement:

 $$\frac{(26^4 * 6^3) * 7!}{ (4!*3!)} = 3,454,738,560 \; combinations.$$

This be the same value as Her Majesty and Sir Cphill

(26^4 * 6^3 * 7!) / (4!*3!) = 3,454,738,560 combinations.

Her Majesty also be correct: “it is the forerunner of a probability question.” One need only add on this: “What be the probability of guessing the key in one try?” You will have to know the above, to answer.

This be why Her Majesty be the Queen and Sir CPhill be the Seigneur: Knight of the manor.

And I be a servant in Her Majesty’s service, who learneth much!


Jul 1, 2014