Assuming that each letter can appear at most once in an arrangement, how many ways are there to arrange 10 letters taken from the alphabet

Question

Assuming that each letter can appear at most once in an arrangement, how many ways are there to arrange 10 letters taken from the alphabet a-z such that: (a) a is not included in the arrangement?(b) z is included in the arrangement? (c) both a and z are included in the arrangement? (d) Exactly one of a and z appears in the arrangement? The intention here is that exactly one element from the set faz) appears in the string. So, abedetohit and bedestohij are OK, but abodedghi is not.(e) z and a appear and are adjacent to each other in the arrangement?

in progress 0
Madelyn 1 week 2021-09-10T15:07:08+00:00 1 Answer 0

Answers ( )

    0
    2021-09-10T15:08:40+00:00

    Answer:

    (a) 25C10 x 10! ways

    (b) 25C9 x 10! ways

    (c) 24C8 x 10! ways

    (d) 2 x 24C9 x 10! ways

    (e) 2 x 24C8 x 9! ways

    Step-by-step explanation:

    Given: We need to arrange 10 letters taken from the alphabet a-z such that:

    (a) a is not included in the arrangement:

       As a is not included in arrangement, we need to select from remaining 25 alphabets. From 25 alphabets, 10 alphabets can be selected in 25C10 ways and can be arranged in 10! ways.

    So, Total no. of ways = 25C10 x 10!

    (b) z is included in the arrangement :

    Now z is included in the arrangement, so we need to select 9 alphabets from 25 alphabets.  

    From 25 alphabets, 9 alphabets can be selected in 25C9 ways and can be arranged in 10! ways.

    So, Total no. of ways = 25C9 x 10!

    (c) both a and z are included in the arrangement:

    In this case, we need to select 8 alphabets from 24 alphabets.  

    From 24 alphabets, 8 alphabets can be selected in 24C8 ways and can be arranged in 10! ways.

    So, Total no. of ways = 24C8 x 10!

    (d) Exactly one of a and z appears in the arrangement:  

    In this case, we need to select one from a and z, so it can be done in 2C1 i.e 2 ways. Now, from 24 alphabets, 9 alphabets can be selected in 24C9 ways and can be arranged in 10! ways.

    So, Total no. of ways = 2 x 24C9 x 10!

    (e) z and a appear and are adjacent to each other in the arrangement
    :

    In this case, we need to select 8 alphabets from 24 alphabets and this can be done in 24C8 ways. z and a can be arranged in 2 ways.

    So, Total no. of ways = 2 x 24C8 x 9!

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )