There are 7893600 permutations.
There are 5*4*3 = 60 permutations.
The number of permutations of the letter ABCDEF is 6 factorial, or 720.
The number of license plates without a repeated letter or digit is 26x25x24x10x9 = 1 404 000. The number of license plates without any restriction is 26x26x26x10x10 = 1 757 600. Probability of a license plate without a repeated letter or digit = 1 404 000 /1 757 600 = 0.7988 Probability of a license plate with a repeated letter or digit = 1-0.7988 = 0.2022
6P4 = 6!/(6-4)! = 6 * 5 * 4 * 3 = 360 four letter permutations from 6 different letters.6C4 = 6!/[4!∙(6-4)!] = 15 four letter combinationsfrom 6 different letters.
26*25 = 650
The number of permutations of the letters EFFECTIVE is 9 factorial or 362,880. Since the letter E is repeated twice we need to divide that by 4, to get 90,720. Since the letter F is repeated once we need to divide that by 2, to get 45,360.
The number of 5 letter arrangements of the letters in the word DANNY is the same as the number of permutations of 5 things taken 5 at a time, which is 120. However, since the letter N is repeated once, the number of distinct permutations is one half of that, or 60.
Six.
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
There are 5*4*3 = 60 permutations.
The Greek letter pi. pi(abcd) represents permutations of the letters in the set {a,b,c,d}.
360. There are 6 letters, so there are 6! (=720) different permutations of 6 letters. However, since the two 'o's are indistinguishable, it is necessary to divide the total number of permutations by the number of permutations of the letter 'o's - 2! = 2 Thus 6! ÷ 2! = 360
The number of ways that the letter of the word CANADA can be arranged is simply the number of permutations of 6 things taken 6 at a time, which is 6 factorial, or 720. However, since the letter A is repeated twice, the number is distinct permutations is a factor of 4 less than that, or 180.
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
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.
Normally, there would be 5!=120 different permutations* of five letters. Since two of the letters are the same, we can each of these permutations will be duplicated once (with the matching letters switched). So there are only half as many, or 60 permutations.* (the correct terminology is "permutation". "combination" means something else.)
An eight-letter word cannot be made from four letters. T, I, E and S are not sufficient to make an eight-letter word without knowing what letters can be repeated and how often they are repeated. TESTIEST could work if you allow three Ts.