NEW PLATES
So the first deployment phase for the new plates had started. This phase's plates are starting ص أ X or س أ x, with the variation of the numbers after that. I have seen cars in Cairo, Mohandessen and even Shoubra (2 cars). All new plates have these variation. I once found 3 cars next to one another with same alpha combination with a different numeric. This I find amuzing!
The second thing is I am now tracing how the translation of the letters from Arabic to english will work. A perfectionist, and detail oriented in trivial stuff I am, I am laughing in sarcasm over the faults they did because of the translation. 'Why did they translate?' I tell my brother. 'How will they account for خ for instance.. and becuase of this translation, the probabilites for plates variation will decrease.' As a big picture guy, he responded 'They have zillion of combinations. Does not matter.' But again, I am now keeping an account how will they use all the letters in translation.
SO far for the problematic letters, here is what they found:
W= و
T= ط
E= ع
G= ج
Y= ي
K= ق
T= ط
So, what about ت؟ خ؟ ث؟ ذ ش ض غ ك ظ ة؟
Probably these numbers will never appear. I wish they put it.
And I keep trying to calculate the probability? LOL
So, it is 26^3 + 25^3 + 24^3? This is for the letters only, right?
ANd then we combine with the numbers. So for every 26^3 we have [9^3 +8^3+ 7^3]....
Am I calculating right!
I will recheck the possibilities. But it is huge I know....
The second thing is I am now tracing how the translation of the letters from Arabic to english will work. A perfectionist, and detail oriented in trivial stuff I am, I am laughing in sarcasm over the faults they did because of the translation. 'Why did they translate?' I tell my brother. 'How will they account for خ for instance.. and becuase of this translation, the probabilites for plates variation will decrease.' As a big picture guy, he responded 'They have zillion of combinations. Does not matter.' But again, I am now keeping an account how will they use all the letters in translation.
SO far for the problematic letters, here is what they found:
W= و
T= ط
E= ع
G= ج
Y= ي
K= ق
T= ط
So, what about ت؟ خ؟ ث؟ ذ ش ض غ ك ظ ة؟
Probably these numbers will never appear. I wish they put it.
And I keep trying to calculate the probability? LOL
So, it is 26^3 + 25^3 + 24^3? This is for the letters only, right?
ANd then we combine with the numbers. So for every 26^3 we have [9^3 +8^3+ 7^3]....
Am I calculating right!
I will recheck the possibilities. But it is huge I know....
Comments
15600 (720)=11232000 combination
or
26*26*26*10*10*10= 1757600 (doesnot make sense) there is no combination
OR
for letters:
26^3+25^3+24^3= 17576+15625+13824=47025
and for every 47025 there is the number combinations
so
1000+729+512=2241
then
47025*2241=105383025 possibilities!
What colour code?
why are you assuming that the letters cant be the same? (hence you are decreasing the options by one each time 26, 25, 24)
Is there an assumption that letters cannot repeat??
if not, then basically it will be 26^3 for the letters only, however my dilemma is how to combine this with the numbers..
thinking of it and will come back :)
1) letters are not mutually exclusive (i.e. can repeat)
2) numbers do not contain zero
now for the letters the possibilities are 26^3 unique entries.. For EACH of these entries, we have 9^3 unique number options they can take..
Therefore: 26^3 * 9^3 = 12,812,904
I feel there is something wrong with this, but I cant find it!
there are 26 letters in the english language and 28 in the arabic, we therefor abide by 26 letters only.
our sample space (s) = 26 elements
but we have to take into account the numbers too, so in this case the sample space is:
26 letters + 10 nos. = 36 symbols
and we only need to find out how many permutations of length 6(3+3)
can be formed from our sample space, which would be equale to the following:
36!/(36-6)! = 1,402,410,240 = 1.4*10^9
so the second deployment of x ط x is taking place.
he adds up 26 letters and 10 numbers to result in 36 symbols.. this assumes that letters are number are equally-weighted, i.e. it assumes that I can have a string of 4 letters and 2 numbers (doesnt enforce the 3 letters and 3 numbers constraint)..
Hence, it is not right :)
I still stick to my proposed solution
but what is the right?!
26!/(26-3)! * 10!/ (10-3)!????
6^V
:)
seriously, this is my first smile since a while! :))
so, no one is able to give the answer.
I was able to tear down all other possibilities but no one tore mine down.
you discredited your own solution by saying that you feel there is something wrong! :)
well, i am not sure.. it does not sound right...
i asked a friend of mine and he came up with same logic..
Again, cant find anything wrong with mine. Am getting obsessed about solving this thing, i dont think THEY put that much effort when they designed it!
Looking back to my high school knowledge there were a thing called Arrangements, abbreviated as A, and also Combination abbreviated as C.
Arrangements shows number of possibilities of arrangements. In order to find Arrangement for 3 items by 3 you do A3=1*2*3=6, ie 123, 132, 213, 231, 312, 321.
Speaking of combination you can take combinations for example among figures 1,2,3 find number of combinations with 2 figures, which is 12, 13, 23, 21, 31, 32, but I don’t remember the formula for that. But I feel that will be an answer, combination of all figures and letters by 6 or 4(depends what was the task). I don’t remember the formula for Combination, but what I remember that combination doesn’t count repetition.
But there could be Arangements of all letters (26) * Arangements of all numbers and then from this amount u take combination of all by 8, ie ABCD1234.
I need to see google and find out how they call this Combination thing correctly, since Combination I mentioned is a translated word, for how we called it in school.
I will come back with an answer later :))))
combinations: ABC = ACB = BAC = BCA = CAB = CBA
i.e. order doesn't matter
permutations: ABC = ABC only.
order does matter.
therefor we are dealing with permutations over here.
nag, you caught me red handed, although it did fly for a while.
after considering this again your solution seems flawless.... but
1.only if the letters are repeated(we can't be sure),
2.you insert the zero back into the numbers (i don't see why you left it out)
the rest seems perfect to me.
if that is the case, then it is the correct solution, haal :p
Nag is a loser.
Annonymous, any concrete input. A la uno.....
She gets an "A" for trying ;)
How many combinations of digits (0-9) are possible for a series of 3 characters (repition allowed)?
Answer: 000 - 999 ... that's 1,000. 10*10*10
Now just for the sake of simplicity, let's suppose we have another 3 characters with exactly the same conditions as above. We also get 1,000 different possibilities.
Combinig both ... for each one of the first, we can have 1,000 of the second. So in total, we have 1,000 * 1,000 possibilities - 1,000,000.
That's 10*10*10*10*10*10 as per haal's self discredited answer in her second comment (although she misses a zero), or (10*10)^3 as per the first anonymous on this list.
Anyone for QED? :):)
I will come back with an answer.
http://en.wikipedia.org/wiki/Combination
So if we want 4 figures and 4 letters it will be
Combination from on top 26 below 4 multiply of Combination from on top 10 by 4, and using the magical link we can find the answer is 3,139,500, hmmmmmm.......
One question, are we having 2 annonymous or just one.
So, can we declare Annonymous 1 (original Annonymous) the winner.
Annonymous 2, Nag and Vile...... Losers.
Vile... first runner
Annonymous 2 Second runner
Nag, as usual the loser. No brussels for you.
Everybody convinced...?
Anyone want to challenge his thought?
Maybe the formula is wrong...
For example you have
3 letter and 3 numbers
Variation of 3 letters is 6, variation of 3 numbers is also 6, which were found by factoring above my me, anonymous (second one), but for each constant combination of 3 letters you have 6 combination of numbers, which in general give you another 6 samples. Therefore total is 6*6 combinations.
Same logic was applied to the 4 letters 4 numbers thing. All combination from 26 letter by 4, multiply by all combination of numbers by 4.
The thing Anon said about 000….999, formula is wrong its like one of the things when you prove 2x2=5, his formula is wrong.
Coming back to your loosers list…. Very generous
Tres momento wont comment anymore. will wait until u come to the right solution, ie MINE.
for the letter: 26. it wont be 1000 actually. it will be 26^3, right?
Then: 26^3 *1000=17576000
right!!
Anonymous 1 (QED guy)
NB. Anon-2, u'll wait for a loooong time :)
But I have one problem... how can we take into account that there will accept repetition in both letters and numbers...one 1 condition, that both repetition at the 2 side wont collide.
for instance, we can have 111 but will have to get different م م م
Any solution?
000/aaa till 999/zzz ... 000-999 is 10^3 cominations ... aaa-zzz is 26^3 combinations.
There is only one (000) just as there is only one (aaa). Don't confuse yourself :):):) Please?
Actually if there is a no-zero rule, she would have come up with the most right solution. Bravo, nag :)
al bravo Nag Al!!!
fyi, she won the award at her company for being the 'problem soluver wizard'. Well, had they asked me, I would have sent her to them long ago. she indeed was misplaced with me.
Any one watching the new plates, they started deploying ط أ x phase recently. They started with ص أ x and س أ x phases.
And yes, I enjoy confusing myself! and everyone else...
If any gets a new plate, please let me know. Interested to see the letter combination. I wont get mine in years. I am not Cairiene.
"she called me to complain about all of you guys and how naiive your solutions were"
you must have me mixed up with someone else :) I wouldve wanted to complain though! it just surprised me that everyone was coming in to say "I know it!! it is combination", whereas you said it was solved through combination the minute you posted the blog, so it was no discovery..
I think what my solution lacked is more confidence from me in it, that's why people were inclined to discard it, because I stated myself that i feel there was sthg wrong with it..
It was a nice mental exercise of stating and defending arguments, and yes, I did lose in that one :D