gate ec

Question

How C Iam getting D

Answer 1

C) is the correct answer. Total 3 digit numbers are 999-99= 900.

Now 21 can be placed at first two places which will give us 10 numbers. When placed at last two places we will get 9 numbers bcoz 0 cannot be placed there. Therefore 900-19 =881

Answer 2

total number of 3 digit numbers = 900
number of digits which contain 21 together = 10+9
total numbers which doesnt contain such pair = 900-19= 881

Comments

please check my answer below and tell me what I missed.

Answer 3

I am getting 882.

Numbers starting with 2 = 2 _ _ = 9*10  (1 cannot be there hence 9)= 90

With 2 as second digit = _2_ = 9*9 (0 cannot MSD and 1 cannot be LSD ) = 81

removing common cases = 22_ = 9

A + B- A intersection B = 162

Now considering other numbers = _ _ _ = 8 * 9 * 10 (0 and 2 cannot be MSD, 2 cannot be second digit as already taken into account) = 720

So totally = 882 (i don't know whether I missed a case)

Comments

Numbers starting with 2 = 2 _ _ = 9*10  (1 cannot be there hence 9)= 90
in this case 221 is possible

---

ya got it thanks :-)