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How many 3 letter permutations of the word October?
There are 195 3-letter permutations.
How many possible permutations using the word bite?
4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------
How many different permutations are there of 3 4 5 6 7?
120
In no order how many 3 digit permutations can be made from 0-9?
"Permutations" do consider the order. If you want the order ignored, then you're looking for "combinations". 3-digit permutations, repetition allowed: 9 x 9 x 9 = 729 3-digit permutations, no repetition: 9 x 8 x 7 = 504 3-digit combinations, no repetition = 504/6 = 84.
How many permutations of 5 voters can be made?
5*4*3*2*1 = 120
How many 3 letter permutations of the word October?
There are 195 3-letter permutations.
How many permutations are there in any 3 from 5?
There are 5P3 = 5!/2! = 5*4*3 = 60 permutations.
How many different 3 letter permutations can be formed from the letters in the word count?
There are 5*4*3 = 60 permutations.
How many permutations are in the word DAD?
because permutations are order specific, and because there are two d's "dad" counts as only one. so it is 6 (3P3)/2 (there are two d's) so the answer is 3 DAD ADD DDA
How many possible permutations using the word bite?
4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------
If you have two football teams in front of two football teams and two football teams behind two football teams and two football teams beside two football teams How many players do you have?
44 players. There would be 4 teams, with 11 players each. Team 1 and 2 would be in front of team 3 and 4 and so would be the two teams in front of two teams, and teams 3 and 4 would be behind teams 1 and 2 and so would be the two teams behind two teams.
How many permutations of 3 different digits are there chosen from the ten digits 0 to 9 inclusive?
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
How many distinct permutations are there of the word statistics?
We know there are 10! (ten factorial) permutations (that's about 3,628,800 permutations); however, we know that number includes repeated permutations, as there are 3 s', 3 t's and 2 i's. So we have to divide by the number of ways these can be written as individual permutations (if they were considered as unique elements), which are 3! (= 6), 3! and 2! (= 2) respectively. So our final calculation would be 3628800 / (6 * 6 * 2) = 50400 unique permutations.
How many permutations are in the word stop?
There are 4*3*2*1 = 24 of them.
How many different permutations are there of 3 4 5 6 7?
120
In no order how many 3 digit permutations can be made from 0-9?
"Permutations" do consider the order. If you want the order ignored, then you're looking for "combinations". 3-digit permutations, repetition allowed: 9 x 9 x 9 = 729 3-digit permutations, no repetition: 9 x 8 x 7 = 504 3-digit combinations, no repetition = 504/6 = 84.
How many ways can 9 books be arranged on a shelf so that 5 of the books are always together?
You can simplify the problem by considering it as two different problems. The first involves consider the five-book chunk as a single book, and calculating the permutations there. The second involves the permutations of the books within the five-book block. Multiplying these together gives you the total permutations. Permutations of five objects is 5!, five gives 5!, so the total permutations are: 5!5! = 5*5*4*4*3*3*2*2 = 263252 = 14,400 permutations
