There are 7893600 permutations.
64 different arrangements are possible.
There are 4 letters in IOWA, all non repeating. Arrangements are 4! or 24.
It depends on whether the order of the numbers is important or not. For example, if 123456 is seen as a different code to 213456 then there are many more possible solutions.If the order is unimportant, the number of possible combinations is equal tobinomial coefficient(15,6) = 5005If the order is important, then the number of possible permutations is equal to15! = ~1.3x1012
There are 12 factorial or 479,001,600 permutations of the letters in the word SOCIOLOGICAL. However, since the letter O occurs three times, the letter I occurs twice, the letter L occurs twice, and the letter C occurs twice, you have to divide that by 8 (O), 2 (I), 2 (L), and 2 (C) giving you 7,484,400 distinctpermutations.
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
2520
There are 7893600 permutations.
The only five letter word that can be made with those letters is 'ditto'.Other words that can be made with the letters in 'ditto' are:dodotIidittotot
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]
The word "spineless" has 9 letters, including 3 s's and 2 e's, so the number of distinct permutations of the letters is: 9!/(3!2!) = 30,240
If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations. If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.
The number of permutations of 26 things taken 4 at a time is 26 factorial minus 4 factorial, or 358,800.By the way, in this day and age, that would be an extremely weak password.
10! permutations of the word "Arithmetic" may be made.
The six letter word that can be made from those letters is global.
No five letter words can be made from those letters. The following words can be made with the letters 'zcmegoy':cogcomecoycozyegogemgymgomemyyo
cheerleader