Yes
Not answereable without knowledge of the data set.
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
The objects within a number set can be caled as "Elements" or "members".
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)! ]
Yes
Yes
The number of permutations of n distinct objects is n! = 1*2*3* ... *n. If a set contains n objects, but k of them are identical (non-distinguishable), then the number of distinct permutations is n!/k!. If the n objects contains j of them of one type, k of another, then there are n!/(j!*k!). The above pattern can be extended. For example, to calculate the number of distinct permutations of the letters of "statistics": Total number of letters: 10 Number of s: 3 Number of t: 3 Number of i: 2 So the answer is 10!/(3!*3!*2!) = 50400
Not answereable without knowledge of the data set.
An arrangement of n objects in a specific order is called a permutation. Permutations refer to all possible ways in which a set of objects can be ordered or arranged.
Permutation Formula A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk . Formula:Example:How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15. different lineups
Using the Permut function, you can find out how many permutations can be got from a set of values. To actually generate the individual permutations you would need a program.
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
You do not have to figure the permutation. You simply rearrange the order of the numbers that you are presented with. The permutations of the number set 1, 2, 3 include 1, 3, 2, and 2, 1, 3.
The Greek letter pi. pi(abcd) represents permutations of the letters in the set {a,b,c,d}.
2.026
The objects within a number set can be caled as "Elements" or "members".