How many $4$ digit numbers can be formed such that they contain the digit $1$ twice ?
My try as follows:
Choose $2$ places out of $4$ for the $2$ ones in $4C2$
Choose $2$ digits out of $9$ for the other $2$ places in $9C2$
Permute the "3-digits" ($2$ ones as one digit , and the other $2$ digits) in $3!$
The answer = $4C2 × 9C2×3!$=$1296$
Is my answer right?
Best Answer
I looked at it this way: the number can start in 2 ways- with a 1 or with other 8 digits ( 0 doesnt count) So for 1 as first:
1*1*9*9+ 1*9*1*9+ 1*9*9*1 =243
for 8 possible digits as first: 8*1*1*9+ 8*1*9*1+ 8*9*1*1 =216
add these two options and you get 243+216 =459