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The number of substrings(of all lengths inclusive) that can be formed from character string of length n is

a)n

b)n^2

c)n(n-1)/2

d)(n(n+1)/2)+1
+1
let's take an example:

s = ABC

length 0 = 1

lenght 1 = {A , B , C} = 3

length 2 = {AB , BC } = 2

length 3 = {ABC} = 1

total string = length 0 + length 1 + length 2 + length  3 = 1 + 3 + 2 + 1 = 7

option D is the correct answer
0
For length 2,why have u not considered AC?
+1
@anji

AC is not  Substring of ABC   because it is separated by B , Substring is that in which we add another substring , in either left are Right side of Substring to get actual Substring.
0
+2

Substring is defined as "Zero or more Consecutive symbols taken"

What you must be confusing it with is "Subsequence" which is defined as "Zero or more Symbols left out (Consecutive or Not)"

Subword is defined as "One or more Consecutive symbols taken" (Subword is same as substring except for Zero length string)

The length of substrings are n or n-1 or n-2 or....or 2 or 1 or 0

let the string is abcd...xyz

no.of substrings of length 'n' = 1           ------  abcd....xyz

no.of substrings of length 'n-1' = 2       ------  abcd....xy , bcd....xyz

no.of substrings of length 'n-2' = 3       ------  abcd....x , bcd....xy , cdef.....xyz

.......

no.of substrings of length '2' = n-1      -----  ab,bc,cd,....vx,xy,yz

no.of substrings of length '1' = n         ------  a,b,c,d,....x,y,z

no.of substrings of length '0' = 1 ----> only empty string is possible

total substrings = 1+2+3+...+(n-1) + n + 1

= [ 1+2+3+...+(n-1) + n ] + 1

= $\frac{n(n+1)}{2}$ + 1

In exam if we just want to find correct option then take any small example..

Like s=ab (n=2)

Substrings=a,b,ab,ε(epsilon)=4 substrings possible.

Hence we found that only last option is correct.:)