Question

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

1.   144
2.   136
3.   120
4.   216

Comments

i am getting 120 as answer but answer given is 136.
my approach :
when B is 0,there is only 1 value of A which satisfies the condition i.e 0 itself.
when B is 1,then there are 2 values OF A - 0000 and 0001
likewise.when B is 15,there are 15 cases of A which are less then B.
hence,1+2+3+.....15 = 120.

Answer 1

Here, 4 digited numbers means the number of binary digits are 4.

Since B >= A, it means we have to count all cases for every possible value of A.

If A = 0000 , B can take values from 0000 to 1111 i.e. 16 values.

A = 0001, B can take 0001 to 1111 i.e. 15 values.

Similarly, it will take 14 , 13 , 12 , 11 , .... 1(when A = 1111 , B can take only 1111).

So, total number of values are 1+2+3+...+16 = 16(16+1)/2 = 136.

So, Option B.

Comments

thanks,i just did calculation mistake.

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Yes, your solution will be valid when B > A (strictly).