The 26 Letters Of The Alphabet Can Make Up How Many Nth Length Word That No Two Consecutive A In A

How many permutations of 4 letters can be made out of the letters of the word 'examination'?

Questions of this type where permutations of a certain length are taken from a word that includes repeated letters are meant to allow repeated letters in the permutations, but identical letters are not to distinguished.For a short example, how many 3 letter permutations taken from 'ball' can be made?  There are 6 with 3 distinct letters—bal, bla, abl, alb, lba, and lab—and there are 6 with 1 letter duplicated—all, lal, lla, bll, lbl, and llb.  That's 12 in all.For four letter permutations of 'examination', we can consider the following cases.All four letters are different.  'examination' has 8 different letters—e, x, a, m, i, n,t , or o—so there are [math]\frac{8!}{4!}=1680[/math] of 4-permutations of these.One letter is duplicated and the other two distinct. Choose 1 of the duplicated letters—a, i, or n.  There are 3 to choose from.  Place them in 2 of the 4 positions.  There are 6 ways of doing that.  Choose 1 of the remaining 7 letters to go in the first empty position, and choose 1 of the remaining 6 letters to go in the last empty position.  That makes [math]3\cdot6\cdot7\cdot6=756[/math] permutations of this typeTwo letters are duplicated.  Choose 2 of the 3 doubled letters.  There are 3 ways to do that. Choose 2 of the 4 positions to place the alphabetically first letter.  There are [math]\binom42=6[/math] ways to do that.  The other letter goes in the other two places.  That makes [math]3\cdot6=18[/math] permutations of this type.Altogether, therefore, there are [math]1680+756+18=2454[/math] permutations in all.

What are all the ten letter country names?

Spaces don’t count as a letter.AzerbaijanBangladeshEl SalvadorGrand DuchyGuadeloupeIsle Of ManIvory CoastKazakhstanKyrgyzstanLuxembourgMadagascarMartiniqueMauritaniaMicronesiaMontenegroMontserratMozambiqueNew ZealandNorth KoreaPuerto RicoSaint KittsSaint LuciaSeychellesSouth KoreaTajikistanUzbekistan

If all possible 10-digit phone numbers could be used, how many phone numbers could exist at the same time?

Let's look at the general case first.If you consider an n-digit phone number, you have 10^n possible phone numbers. Why?If you have a one-digit number, you have the numbers 0 through to 9, which is ten possible numbersIf you have a two-digit phone number, the first digit has possible values 0 through to 9; for each of those you have a second digit, which is also 0 through to 9, giving you ten times ten possible numbers (a hundred possible numbers).If you have a three-digit phone number, the first and second digits have 100 possible values; for each of those the third digit has 10 possible values, giving 100*10 = 1000 possible numbers.ETC So once we get to the 10th digit, we have 10^10 possible numbers. How much is 10^10? It's 10'000'000'000, which many people refer to as 10 billion.

How do I increment each character in a string using Java?

1.Create a new String .2. Use str.charAt(index) method of String class to access each character.3.And simply add ‘1’ to it and type cast it back to char4.And add it back to String.As explained in below code:String str="ABCDEF";
String strIncremented=new String();
for(int i=0;i strIncremented+=(char)(str.charAt(i)+1);
}
System.out.println(strIncremented);
But it’s time complexity is O(n^2) instead of O(n) where n is the length of the string. Reason being strings are immutable in java. So ,when you append a new character to a string instead of just adding it to the end of string ,it copies the whole previous length of string and then appneds the last character.So ,what to do ??Use StringBuilder :StringBuilder sb=new StringBuilder();
for(int i=0;i sb.append((char)(str.charAt(i)+1));
}
System.out.println(sb);
With StringBuilder the time complexity is O(n) as it just appends to the end of the string. :)