There are 195 3-letter permutations.
4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------
120
"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.
5*4*3*2*1 = 120
There are 195 3-letter permutations.
There are 5P3 = 5!/2! = 5*4*3 = 60 permutations.
There are 5*4*3 = 60 permutations.
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
4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------
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?
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.
There are 4*3*2*1 = 24 of them.
120
"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.
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