# Can this be solved with a mathmatical Equation?

I need to figure out how many differant possibilities there are in any given sequence of lettters areanged with multiple letters in each sequence. Example: We have the letters ABCDE and arrange them in one letter sequences.

A

B

C

D

E

There are 5 differant possibilities. Now if we arrange them in 2 letter sequences

A

AA

AB

AC

Ect.

Anyway is there a mathamtical equation that can tell me how many differant possibilities there are with the length of the sequence and the ammount of possible characters in the sequence as variables?

### 3 Answers

- PuzzlingLv 71 decade agoFavorite Answer
Yes, assuming you can have repeats in the letters:

For one letter you have 5 choices of A, B, C, D or E.

= 5

{A, B, C, D, E}

For two letters you have 5 choices for the first position and 5 choices for the second position:

= 5 x 5

= 5²

= 25

{AA, AB, AC, AD, AE, BA, BB, BC, BD, BE, CA, CB, CC, CD, CE, DA, DB, DC, DD, DE, EA, EB, EC, ED, EE}

For three letters you have 5 choices for the first position, 5 choices for the second position and 5 choices for the third position:

= 5 x 5 x 5

= 5^3

= 125

I won't enumerate them but it would be the same set as above, but follow each one with A,B,C,D,E:

{AAA, AAB, AAC, AAD, AAE, ABA, ABB, ABC, ABD, ABE, ACA, ACB, ACC, ACD, ACE, ADA, ADB, ADC, ADD, ADE, AEA, AEB, AEC, AED, AEE, BAA, BAB, etc.}

For four letters:

= 5 x 5 x 5 x 5

= 5^4

= 625

For five letters

= 5 x 5 x 5 x 5 x 5

= 3125

In general, for 5 items of length n (with repeats possible) the number of arrangements is:

= 5^n

Finally if you just want the ways of creating 1, 2, 3, 4 or 5 letter arrangments, add them all up:

5 + 25 + 125 + 625 + 3125

= 3,905

- Anonymous4 years ago
Multiply out the brackets: 3x - 12 = 5 - x upload x to the two area 4x - 12 = 5 - x + x upload 12 to the two area 4x - 12 + 12 = 17 divide the two area by using 4 x = 17 / 4 or 4 (it relatively is sixteen / 4) and a million/4 (it relatively is .25) x = 4.25