counting

Question

Let A,B are 4 digited numbers, how many possible cases are there to have B≥A?

Answer 1

Let A = 1000, So B=1000 to 9999 will satisfy i,e. 9999-1000+1 = 9000 cases

now A = 1001, So B=1001 to 9999 will satisfy i,.e 8999 cases

now A=1002, So B=1002 to 9999 will satisfy i,.e 8998 cases

similarly

..

..

..

now A= 9999, So B=9999 will satsfy i,e. 1 case

So, total cases = 9000 + 8999 + 8998 + 8997 + ..... 1

=9000(9001)/2=40504500

Comments

this is given in solution which i dint get

Number of total terms with n digits =2ⁿ
Number of equal n digited numbers (A=B)=2ⁿ
Number of cases (A>B)=2^(2n-1) - 2^(n-1)
Number of cases (B>A)= 2^(2n-1) - 2^(n-1)
Number of cases (B≥A)=2^(2n-1) - 2^(n-1) + 2ⁿ
Here n=4
⟹2^(2×4-1) - 2^(4-1) + 2⁴
=2⁷ - 2³ + 2⁴=128 - 8 + 16 = 136

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You have taken the digits in binary number system.

if A=0000 then B will be 0000 to 1111 i,.e 16 cases

if A=0001 then 15 cases and so on

16+15+14+13+......1

=n×(n+1)/2=16×17/2=136

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your solution is based on the concept of comparator.