Sir doubt in this question

#1

sir what is the reason for using this formula
(j-i)*(j-i+1)/2
what is the reason behind this??

#2

hello @tejasddongare
let say we have length L subarray with all elements distinct.
then all subarray formed from this L length subarray will also contain distinct elements right?

so how many subarray possible from this L length subarray .
it will be LC2 + L -> L*(L-1)/2+ L
= > L*(L+1)/2
if L = j-i
then
=> (j-i) *(j-i+1)/2

#3

sir i dont know permutation and combination can you show me different mathematical proof

#4

ok , so from L lenght subarray
how many 1 lenght subarray possible -> L (right?)
how many 2 lenght subarray possible -> L-1
how many 3 lenght subarray possible-> L-2
how many 4 lenght subarray possible-> L-3
…
how many L lenght subarray possible-> 1

now compute its sum-> L+ L-1 + L-2 + L-3 + L-4 …+ 2 + 1
it is simple summation -> L*(L+1)/2

#5

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