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
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!