The number of permutations of the letters MASS where S needs to be the first letter is the same as the number of permutations of the letters MAS, which is 3 factorial, or 6. SMAS SMSA SAMS SASM SSMA SSAM
NINON
The answer would be 210
The first letter can be any one of 26. For each of these ...The second letter can be any one of the remaining 25. For each of these ...The third letter can be any one of the remaining 24.So the number of different 3-letter line-ups is (26! / 23!) = (26 x 25 x 24) = 15,600.That's the answer if you care about the sequence of the letters, i.e. if you call ABC and ACB different.If you don't care about the order of the 3 letters ... if ABC, ACB, BAC, BCA, CAB, and CBA are allthe same to you, then there are six ways to arrange each group of 3 different letters.Then the total number of different picks is [ 26! / (23! 6!) ] = (15,600/6) = 2,600 .
There are thirty letters in the German alphabet. The first 26 are the same as the English alphabet, however they may be pronounced different. There are four extra letters, one of which is called the Esstsett.
If you mean re-arranging the letters 'g', 'r', 'e', 'a', and 't' in different sequences, then: We have five letters. The first letter can be any one of the five.For each choice, the second letter can be any one of the remaining four.For each choice, the third letter can be any one of the remaining three.For each choice, the fourth letter can be either one of the remaining two.Then there is only one letter left to put in the fifth place. So there are (5 x 4 x 3 x 2 x 1) = 120 different ways to carry out this process.
Wheat
The first letter can be any one of 26 letters. For each of those . . .The second letter can be any one of the remaining 25 letters.There are (26 x 25) = 650 different 2-letter codes.
There are 6 different letters. That means you have 6 choices for placing the first letter. Once you have placed that letter, you now have 5 choices for the second letter. So you have 6*5 = 30 options for the first two letters. Continuing this way, you have 6*5*4*3*2*1 = 720 ways to rearrange the 6 letters.
The maximum number of different three letter palindromes that can be made using any letter of the alphabet is 1 - that letter repeated 3 times. However, if you meant you can pick any letter for each letter of the palindrome, then: The first and last letters are the same and there is a choice of 26 letters for them; For each choice for the first (and last) letter, there is a choice of 26 letters for the second letter Making a total of 26 × 26 = 676 such three letter palindromes (where case of the letter does not matter).
Different species have different names that start with different letters. However, all species names are generally preceded by a genus name.
The first letter can be any one of the 6 letters. For each of those . . .The second letter can be any one of the remaining 5 letters. For each of those . . .The third letter can be any one of the remaining 4 letters. For each of those . . .The fourth letter can be any one of the remaining 3 letters. For each of those . . .The fifth letter can be any one of the remaining 2 letters.The total number of different ways they can be arranged is (6 x 5 x 4 x 3 x 2) = 720 .
Some seven letter words with the first and sixth letters the same are:abysmalaccrualacrobatadmiralalmanacanagramangulararrivalartisanarsenalcapricechalicecollectcompactconcoctconductconnectcontactconvictcorrectcornicecrevicedecidedegradeearlierepithetescapeefalsifyfortifygarbagegeologyhealthyidioticillicitimperilinertiaingraininhabitinheritinhibitinsulininvalidkaraokelegallylegiblylightlylikablelocallylooselylovableloyallyluckilymacramenirvananothingoctagonopinionoutlookovationoxbloodperhapspreceptpreemptrampartrapturerectoryrequirerestorerevelryrivalryrhubarbrupturesadnesssarcasmselfishsoloiststylishsubsistsuccesssucrosesuggestsupposesurmisesurpasssynapsetaffetatermitethirstythriftytributetrinityuraniumwillowy
The number of arrangements of the letters PARTY, if the first letter must be an A is the same as the number of arrangements of the letters PRTY, and that is 4 factorial, or 24.
There are 26 letters in the alphabet.The first letter is A and it ends with Z.
adele
The first letter of bandits password is M and his pass is four letters!
Assuming you don't repeat letters:* 7 options for the first letter * 6 options for the second letter * 5 options for the third letter * 4 options for the fourth letter (Multiply all of the above together.)