A keycode must contain 4 letters and 3 numbers. The letters may be any letter of the alphabet. The numbers should be any number from 0 to 5. How many different keycode combinations are there?

Guest Jul 1, 2014

#1**+14 **

Well, If the numbers all go first it will be 6^3*26^4

Now if they are all mixed up you are testing me. Probability is a bit of a joke on this forum.

Now let me see. there are 7 items in the code.

I'm going for 7C3*6^3*26^4 =

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{26}}}^{{\mathtt{4}}} = {\mathtt{3\,454\,738\,560}}$$

It is probably wrong. I've got other things to do. See you later!

Melody Jul 1, 2014

#2**+5 **

We can choose any 4 of 26 letters

And we can choose any 3 of 6 numbers

And we have 7! arrangements of these letters and numbers

So we have

C(26,4)* C(6,3) *7! = 1,506,960,000 combinations

CPhill Jul 1, 2014

#4**0 **

Well.....I don't know if my answer is * actually* correct, zegroes.....it's just my "take" on the problem!!!

CPhill Jul 1, 2014

#5**0 **

Yes ......... maybe........but they are not alive though!

But CPhill is one great and unique fossil whose alive !or maybe its just his spirit!lol!

rosala Jul 1, 2014

#7**+5 **

You know Chris we have interpreted the question differently from each other. Mine is not correct.

I interpeted this as all the letters could be the same and all the numbers could be the same.

the 7C3 part of mine is not correct because the letters could all be the same. So my number will be too high.

-------------------------------------

Now let me think about yours.

**Firstly you have a simpler problem than mine because you has assumed that all the letters and all the numbers have to be different from each other. So yes, based on this interpretation I do believe yours is correct.**

Melody Jul 1, 2014

#8**0 **

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}

MorganTud Jul 1, 2014

#10**+3 **

Well Mr Tud, You are correct - this in itself is not a probablity question but it is the forerunner of a probablility question so I have very loosely categorised it as such.

You have chosen the 3 numbers and the 4 letters just as I did. however, this is a key code. It is reasonable to assume that order does count so I am sorry Chiurgeon Tud but your answer is not correct!

Melody Jul 1, 2014

#11**+3 **

It may look simple Zegroes but it is not correct.

Given Chris's interpretation - His answer is the only correct one.

Melody Jul 1, 2014

#13**0 **

**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}

MorganTud Jul 1, 2014

#14**+5 **

I was assumig that all the letters and numbers were different......this is possibly not correct (but it could be)

If we can have "repeats" then I believe my answer should be:

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

Which kinda' looks just like Melody"s !!!!

Are we mistaken???? I don't know !!!!

CPhill Jul 1, 2014

#15**0 **

No Chris the 7C3 bit does not work if some of the digits/letters are the same.

The solution becomes much more complicated. I might have another crack at it in the morning - I am much too tired now.

Good night all.

Melody Jul 1, 2014

#17**0 **

Zegroes instead of this why dont u think of that question which says"8over7devided by56"!theres one battle of answers going on in there also in which u too are involved!(or maybe u r be get out in the battle)lol!

rosala Jul 1, 2014

#19**0 **

maybe...........but ur formula was a lil "upside down"kind of !and mine was easier to understand so i get more points !

rosala Jul 1, 2014

#20**+11 **

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 }

MorganTud Jul 1, 2014

#21**0 **

I've been giving some thought to this one....I believe Melody is correct.....our previous answers were incorrect with respect to allowing repeats.....I think that the answer is * WAY *larger.......

I see this as a "slot" problem with these considerations........

1. Choose any 4 of of 7 "slots" and let each slot be occupied by any one of 26 letters. Notice that order* isn't *a consideration, since, for example, "slots" 1-2-4-7 are the same as 2-1-4-7 or 1-4-2-7.

2. For each arrangement above, the other 3 "slots" can be occupied by any three digit combination utilizing numbers 0 through 5.......there are 556 of these possible combinations.......(000 through 555).

So we have:

C(7,4)* (4^26) * 556 = 87,640,048,748,629,852,160 combinations !!!

Whew!!!....that's a pretty big number!!! (Note that the 4^26 part contributes to most of the "bigness".....)

Anyone else have any ideas on this??

CPhill Jul 2, 2014

#22**0 **

Nope I don't think so. At this stage I think this is a massively long and convolured problem so I am just working on it in bits. I'd really like someone to point out the error in my logic below. Thank you.

I've got a new train of thought.

This is just a little bit of my newest thought pattern

I am just looking at number possibilities of the 4 letters here.

4 letters are chosen from the alphabet. And these are supposed to be the number of possibilities but something is wrong!

Letters all different = 26*25*24*23=358800

letters 2 the same= 26*1*25*24*6= 93600

letters 3 the same=26*1*1*25*4=2600

Letters all the same=26

Add all these and I get 455026

Now I know it should add up to 26^{4}=456976 so I am short a few possibilities here.

**WHY AM I SHORT???**

NOTE: 456976-455026=1950=25*26*3 (does this help?)

Melody Jul 5, 2014

#29**0 **

I want to simplify this question and play with it.

We have 26 letters and 6 digits . Repetition is allowed.

I want to know how many combinations of 2 letters and 2 numbers that there are.

-------------------

there are 26^2=676 letter combinations

Now I know that in 26*1=26 of these the 2 letters are the same so the rest must be different.

That would be 26*25= 650 (Yes that works)

-------------------

there are 6^2=36 number combinations

6 have repetition and 30 do not

-------------------

Now when I put this together i can have

2 letters the same and 2 numbers the same 4!/(2!2!) = 3

26*6*3=468

2 letters the same and 2 numbers different 4!/2! = 12

26*30*12= 9360

2 letters diff and 2 numbers the same 4!/2!=12

26*25*6*12= 46800

2 letters diff and 2 numbers diff 4! = 24

26*25*6*5*24 = 78000

SO I think that the number of 4 tumbler codes would be 468+9360+46800+78000= 134,628

--------------------------------------------------------------------

NOW, I would like Sir CPhill and Chiurgeon Tud to comment on my solution to this simplified problem please.

Do you think that it is correct?

Melody Jul 6, 2014

#30**+3 **

I just had a thought!

The letters and numbers can be ordered when they are first collected or when they are fitted to the tumbler.

I think Tud (and therefore the original Melody and CPhill) might be right afterall

If the 3 numbers are collected such that order matters then there will be 6^3 possibilities

If the 4 numbers are collected such that order matters then there will be 26^4 possibilites.

Since order is already taken into account then now we just need to know how many ways 3 can be chosen from 7 without order 7C3

So yes for the moment I am thinking that our original answers were correct and i can understand Tud's frustration at my inability to see that

So for now i am going with

$$6^3*26^4*7C3 = 3,454,738,560$$

I reserve the right to change my mind again!

Now i have to think about whether the post above is correct or not. It probably is not. I will think about it again another time.

Melody Jul 6, 2014

#34**0 **

New Info (with Bertie's and Rom's help- thanks guys.)

4 letters are chosen from the alphabet. And these are the number of possibilities.

Letters all different = 26*25*24*23=358800

letters 2 the same and other 2 different= 26*1*25*24***4C2**= 93600

letters 2 pairs = 26*1*25*1*4C2 divided by a half (got it) = 1950

I had to halve it because for each pair left there are 2 that are the same.

letters 3 the same=26*1*1*25***4C3**=2600

Letters all the same=26

Now I know it should add up to 26^{4}=456976 and they do so that is great!

-----------------------------------------------------------------

I might use this to figure out something else later. Mmm not sure!

Melody Jul 9, 2014

#41**+14 **

Moonlight at the Taj Mahal staring Rosala and Zergroes.

... Stay tuned for more intrigue.

SevenUP Jul 9, 2014

#42**+11 **

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 }

MorganTud Jul 9, 2014

#43**0 **

I see CPhill also suffers the inversion affliction.

Chiurgeon Tud, hath thou tested thineself for the dyslexia condition. It may be prudent to do so. 363 could hence become 633 tis a code that some may quite enjoy.

Well, I bid thee a good night/day Chiurgeon Tud.

Lady Guinevere.

Queen of Camelot

Melody Jul 9, 2014

#44**+9 **

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}

MorganTud Jul 9, 2014

#47**0 **

Zegroes this is really a very old post and ur still scrolling over those "century"old posts!thats funny!

rosala Jul 9, 2014

#48**0 **

Actually if you look at morgan tuds last post above my previous one you will see its ONLY 2 HOURS

zegroes Jul 9, 2014