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5! (Factorial).
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When you make an organized list to find the permutations of the letters in the word BITE how many permutations are English words?
Only one.
How many permutations are there of the word?
The word would be needed for a specific response, but here's a general one: Take the factorial of the number of letters in the word, and divide that by the product of the factorials of the counts of all the unique letters in the word. An example: For the word "levitation", there are 10 letters, with one L, one E, one V, two Is, two Ts, one A, one O, and one N. So the solution is: 10!/(1!1!1!2!2!1!1!1!) = 10!/4 = 907,200 permutations
How many different 4 letter permutations are there for the letters in the word toolroom?
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)! ]
How many ways can you make 8 using 3 dices?
There are 18 permutations or 5 combinations. And it is one die, many dice. There is no such word as dices or diceses.
What are the number of permutations of the letters of the word MATH?
For the first letter, you can choose any of the four. For the second letter, you can choose any of the remaining three. For the third letter, you can choose either of the remaining two. For the fourth letter, you don't get a choice, you have to use the one that's left. So 4x3x2x1 = 24 possible permutations.
When you make an organized list to find the permutations of the letters in the word BITE how many permutations are English words?
Only one.
How many arrangements of the letters in the word SCHOOLS are there?
The number of permutations of the letters in the word SCHOOLS is the number of permutations of 7 things taken 7 at a time, which is 5040. However, since two of the letters, S and O, are duplicated, the number of distinct permutations is one fourth of that, or 1260.
How many permutations are there in a 3-letter code which can use the letters from A to Z inclusive?
A - Z means you can use the whole alphabet, which usually contains 26 letters. So a one-letter code would give you 26 permutations. 2 letters will give you 26 x 26 permutations. A three letter code, finally, will give you 26 x 26 x 26 , provided you don't have any restrictions given, like avoiding codes formed from 3 similar letters and such.
How many different 5 letter arrangements can be formed from the letters in the word Danny?
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.
How many permutations are there of the word?
The word would be needed for a specific response, but here's a general one: Take the factorial of the number of letters in the word, and divide that by the product of the factorials of the counts of all the unique letters in the word. An example: For the word "levitation", there are 10 letters, with one L, one E, one V, two Is, two Ts, one A, one O, and one N. So the solution is: 10!/(1!1!1!2!2!1!1!1!) = 10!/4 = 907,200 permutations
How many ways the word probability can arrange?
There are 11 factorial, or 39,916,800 permutations of the letters in the word PROBABILITY. However, because the letters B and I are duplicated, there are only one quarter of that, or 9,979,200 distinguishable permutations.
How many different 4 letter permutations are there for the letters in the word toolroom?
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)! ]
How many ways can you make 8 using 3 dices?
There are 18 permutations or 5 combinations. And it is one die, many dice. There is no such word as dices or diceses.
How many words can you make using all the letters in computer at one time?
You can make 40,320 words from the letters in computer
How many ways can the letters in the word mississippi be rearranged if the two Ps must stay together?
Rethink the question as "How many ways can the letters of MISSISSIPI (one P) be arranged?" The (initial) answer is that this is the number of permutations of 10 things taken 10 at a time, because every time you choose a P, you simply write down two P's. That is 10 factorial, which is 3,628,800. However, since there are still multiple letters (four S's, and four I's), you need to divide by 24 twice in order to see how many distinct permutations there are. That is 3,628,800 / 16 / 16, or 14,175.
How many combinations are there in a seven digit phone number?
There is only one combination. There are many permutations, though.
How many words can you make out of the letters of One Hundred Days?
hundred
