# 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?

Relevance

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:

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

• Anonymous
4 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

• 5^n where n is the length of sequence.