How can i arrange 14 letters in various orders without repetition?

Question

how can i arrange 14 objects in various different orders without repetition.?
if i have 14 letters i.e. A,B,....LMN. with 3 replications i.e A1,A2,A3 .....N1,N2,N3. can you please arrange these letters for me in various different orders without repetition of a particular letter?e.g First order =A1,B3,C1,D2,E2,F1,G3,H1,...
Second order=A2,B3,C1,D3,E3,F1,G...

I will be very grateful because i don`t know the permutation concept or any software that can do this.Thanks in advance.my email is [e-mail]

Answer 1

Since you have 52 different letters and you don't need repetitions then you are asking for Permutation. Take 51 of the 52 letters and the number of permutations possible = 52!/(52-51)!x51!. This much permutations is possible. The remaining one you can do which is missing. Probably this is a programming project. You can programme this formula, generate the different permutations and go from there.

Answer 2

This answer does not give you the arrangements but gives you the total number of arrangements.

We use the permutation formula
n! ( n factorial) and in this case n = 14

n! = n X n-1 X n-2 X n-3 X n-4 X......... X 1

So 14! = 14 X 13 X 12 X 11 X 10 X 9 X 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 87,178,291,200. This is the number of times 14 objects would be arranged without repetition

Yahoo Answers cannot take a full list of these permutations.

When taking 3 at a time,

we have 14P3 (where 3 is written as a subscript)

14P3 = 14 X 13 X 12 = 2184.

This is the number 14 different objects, with 3 replications at a time

Answer 3

Let me try again, I misunderstood the first time.

You have 14 slots to fill in. The first slot can be any of 42 (= 3*14) different values. The second can be any of 39 different values: it cannot be the first one chosen, or any of the other copies of it.

Continuing, the number you are looking for is 42*39*36...*6*3 (or, 14!*3^14)

Or, if you prefer, consider assigning the letters and the replications separately. First order the 14 letters, there are 14! ways of doing this. Next go back to the slots and add the replication. In each case you can choose from {1, 2, 3}, so there are 3 independent choices in each of the 14 slots, so multiplying this also gives 14!*3^14.

Answer 4

A1 can be arranged with B1,B2,B3. same with A2 and A3. So we can arrange AnBn in nine ways . Similarly if you put C1 then they can be arranged in 3^3 ways. so 14 letters can be arranged in 3^14 ways. Now if it is so that ABC doesn't have to be in order then . ABC...LMN can be arranged in 14! ways, so the no. of ways will be 3^14*14!.

Answer 5

First, evaluate how many approaches there are to reserve all 4 letters. it somewhat is a controversy of diversifications. all of us understand that the form of diversifications of n aspects is n!. as a result, n=4, so there are 4 × 3 × 2 × a million = 24 diversifications. despite if, we in ordinary terms care approximately how the 1st 2 aspects are ordered. we are distinctive-counting by way of the form of approaches the 0.33 and fourth aspects could be ordered. for example, RSTU and RSUT the two supply us RS as a results of fact the ordered pair. we could desire to divide the form of diversifications of four aspects by way of the form of diversifications of two aspects with a view to do away with the distinctive counting. this is basic. The form of diversifications of two aspects is two! = 2 × a million = 2. So the form of ordered pairs we are in a position to make out of the climate R, S, T, and U is (4 × 3 × 2) / 2 = 24 / 2 = 12.

Answer 6

we had 14 letters repeating them 3 times each
i.e, we get total 14*3=52 letters
we cn arrange them in 52! number of ways