MorganTud

Greetings Your Majesty,

Please forgive the lateness of thy servant’s reply.

The crystal sphere often standeth at the beck and call, and to my commands it harkens and obeyeth, except when my voice travels through a bottle where there be waters of fire. Indeed, the crystal sphere will then taketh me to strange places: They be such as the private harems of mighty sultans, and the planet of the war god, Mars, where I did watch the transit of Earth cross the sun. These be always and forever a wonder to behold, and this be so, no matter how much rum and plaid waters of fire circle in my veins.

Indeed, your Majesty, there be many flavors of this CDD where I be. Here it be known as Contagious Darkness Disease. This be the Dark Ages, and most be born into this curse. There be children of lesser gods who seeketh to escape and they be dragged back in by the gods of their fathers.

There doeth seem to be hope, but only for they who diligently seeketh to escape the clutches of this devil. To they who listen, I sayeth as one who be a practitioner of the medical arts, avoid those who embraceth this as a way of life, for it be real and true as a way unto death of mind and body, and they who maketh a long acquaintance of the affected shall too find themselves affected with this darkness.

In your Majesty’s service

Morgan Tud _M^-1

Greetings to you, Sir CPhill:

I seeth ye have made changes to thy avatar and address. The byway of 666 hath and leadeth to many remarkable wizards and sorcerers. Yea, it passeth by the castle and mansions of Morgan le Fay, who be one with the greats and who resideth near that byway.

The path be potent and hath may powers. The bolder of Sisyphus hath rolled on this path in times of times, and ye may even findeth the Roman ZERO for which ye be searching.

By your leave,

M^-1

Greetings Your Majesty:

Here be a possible reason why the image ye seek come not forthwith.

There be many incantations in the crystal sphere that call forth spirits and graven images made by spirits.

In the beginning, the surf who made the post calleth the images from many domains and the images came forth. Now, it be that one of the domains hath vanished. Some of these domains be temporary and appeareth for a little while then vanishes away.

After the domain vanishes, if wizard’s assistant in the crystal sphere calleth for a graven image the wizard findeth that it does not exist, and never will again in this place, then to saveth time, the incantation be removed, for there now be no reason to use an incantation to call forth a graven image from a domain that exists not.

The graven images for choices of answers be in domains that still exist, and the incantation to call them still exists, so, they display when called.

In Her Majesty’s service,

M^-1

Yea, Your Majesy. This be why I divide by the factorial of redundancy. This do treat for certain dyslexia conditions.

In Your service,

M^-1

What be this impudent consideration that yon commoners attachéth to the posts of Her Majesty and Sir Chpill?

Aye! I see Her Highness, Princess Kitty hath graced us with a visit. Welcome to the keycode, Your Highness. In your service, as always.

Morgan Tud M^-1

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!

M^-1

**it reasonable to assume that order does count**

Aye, my Lady, Thee be true, order doeth count.

However, all orders be represented here with repeatable icons. For the beginning may be AAAA000 and the ending may be 555ZZZZ. Commutating these or any arrangement would be included in the sequence.

I standeth by my answer.

Her Majesty’s solution to this be correct.

I no longer standeth by my answer.

The explaination be forthwith appended.

In your service

M^-1

This be not a probability question.

This be a sequence from a repeatable selection of icons.

This be simply

$$26^4 * 6^3 = \ 98,706,816$$

(no need to commutate as this produces indiscernible redundancies)

Her Majesty’s solution to this be correct.

There be a need to permutate.

The explaination be forthwith appended.

In your service

M^-1

Aye, Sir Cphill, this hath the sound of great power, I shall investigate in this inst. this not(quadratic) magick of which ye speak.

Doeth not the Magick of the Quadratic findeth the Zeros?

For now, there be no less than three kittys that needeth my attention. Appeared suddenly they did, not sure from thither they come.

By your leave,

M^-1

Aye! But the power of Majesty maketh that which be less than, now be greater than.

M^-1