Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
5040
There are 8! = 40320 permutations.
There are 6! = 720 permutations.
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
There are 195 3-letter permutations.
Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
There are 7893600 permutations.
4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
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)! ]
There are 5*4*3 = 60 permutations.
12
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Imagine you have four empty buckets in which to put any of the four letters Y, A, R, D into. At first you have four letters that can be placed in any of the four empty buckets. Once you've placed a letter in a bucket you only have three letters and three empty buckets to choose from. And so on... So there are 4x3x2x1 = 24 permutations of the word YARD.
loom, loot, molt, mool, moor, moot, mort, room, root, rotl, roto, tool, toom, toro
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.