GATE ECE 2017 Set 2 | GA | Question: 9

Question

The number of $3$-digit numbers such that the digit $1$ is never to the immediate right of $2$ is

• A. $781$
• B. $791$
• C. $881$
• D. $891$

Answer 1

Answer C

The total number of three digit numbers is $9\times 10\times 10$ as the first place can be filled with $9$ ways $(1-9)$ and next two places can be filled with $10$ digits $(0-9)$ in $10 \times 10$ ways.

Now the total number of three digit number with $1$ immediate right of $2$ are

• $(210-219) \implies 10$
• $ \_$   $2$ $1 \implies$ this blank can be filled from $1-9 $ so total $ 9$ numbers here
• Thus $10 +9 = 19$ numbers are having $1 $ to the immediate right of $2.$

So, total number of three digit numbers not having $1$ to the immediate right of $2 = 900 - 19 = 881$.

Answer 2

Because 0 cannot come in the first place then 9 ways to fill first position, 10 ways for the next position and same way 10 ways for next position.

9*10*10=900 ways to have 3 digit numbers which dont start with 0

next there are 10 ways where first position is 1^st  and 2^nd position numbers are 1 and 2(121,122,123,124,125,126,127,128,129) since this is the only way to get position 1 and position 2 1 and 2 respectively then position 2 and 3 we can get in 9 ways 112,212,312,412,512,612,712,812,912 so sum of that is 19 so 900-19=881 ways.

Answer 3

Total Possible 3 digit number that can be formed is 900 (since we can’t place 0 in front so 9 * 10 * 10).

now given condition that 2 is in right of 1 can be satisfied if number are placed in  (x 2 1) or  (2 1 x) order. so since 9 numbers can be placed in former order and 10 in later. so answer = 900 – 10 – 9 = 881. Option C is correct.

Answer 4

https://drive.google.com/file/d/1tY8Qrc5BVzidtcEdcrpoG2TxkZPd2tED/view?usp=drivesdk   follow this  for detail solution

Comments

could have just pasted it here