Aug 9, 2010 · looking at the alphabet ,the letters are numbered 1-26 , such that 1 =one=15+14+5=34 (O=15, N=14, E =5 ) 2=two=20+23+15=58 (T=20, W=23, 0=15) 3=three =56 4=four=6...

Jul 27, 2015 · I'm wondering how many possible 3 character combinations can be made using the 26 letters of the alphabet, and 0-9. I've seen that with just the alphabet you can create somewhere around 17k different combinations of 3 letters, but I assume adding in 0-9 must increase that number substantially.

If the letters are randomly assigned, a search for three numbers and a letter will find on average $\frac {74}{26} \approx 3$ matches, but there will be dispersion on that. To return on average $11$ matches, you could search for three numbers and a letter, require exactly those numbers, but accept letters up to three ahead (going around the ...

Aug 24, 2015 · We get aa, ab,ac, and so on up to az. So there are $26$ words on the first line. On the second line, list all the words that begin with b, so ba, bb, bc, and so on. The second line has $26$ words. On the third line, list the $26$ words that begin with c. And so on. We have $26$ lines, each of which has $26$ words, for a total of $(26)(26 ...

26 lowercase letters (a-z) 26 uppercase letters (A-Z) 10 digits (0-9) 33 punctuations and special characters; How many total choices can each character within the password use? ADD the above numbers to answer that: $26+26+10+33 = 95$ How many characters is the password in question? I believe we identified 8 in this scenario.

Apr 19, 2016 · Case 1 – Exactly one number. Case 1a – The letter is in position 7 and the special character is in 8 i.e. LLLLLL9* so, there are 26^6 x 10 x 9 possibilities. Back to the general case 1, how many other arrangements are there? The digit can be in any of 8 positions and the special character in any of the other 7. So 8 x 7 x the number of 1a ...

Feb 17, 2015 · What is the number of ways to order the 26 letters of alphabet so that no two of the vowels a,e,i,o,u occur consecutively? What I thought is to subtract permutations consisting of 2 vowels, 3 vowels, 4 vowels and 5 vowels occurring consecutively from all permutations 26!. But I could not even find an reasonable answer with that method.

How many five-letter palindromes are there (using ordinary 26 letter alphabet)? I was thinking it was either using permutation with first three leters or using 26 cubed. Thanks

If all characters are letters ($26^4$ possibilities) or all characters are numbers ($10^4$ possibilities, none of which doubles up with a case where all characters are letters) then our string is not a password. In all other cases the string IS a password. Hence there are $$36^4-(26^4+10^4)=1\ 212\ 640$$ possible passwords.

Apr 30, 2014 · Another way to see this is to see that the probability that A comes before B is the same as the probability that A comes before some other letter. There are 25 letters that are not A. But there is the case where A is the last letter. There are 26 cases then, each having the same number of outcomes. The probability of each is 1/26.