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

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  • 1 decade ago
    Favorite 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

  • 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

  • 1 decade ago

    5^n where n is the length of sequence.

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