# How many 4 digit numbers can be formed such that they contain the digit 1 twice?

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?

• It depends on whether 0112 is a valid 4 digit number. – Glorfindel Apr 28 '17 at 8:08
• @Glorfindel absolutely invalid ; but how can i remove such cases? – user373141 Apr 28 '17 at 8:10
• No, that's not the way to go. You could have guessed that the answer is wrong: there are 10000 (or 9000, depending on the definition) possible 4 digit numbers; 1296 is more than 10% of that, which is way too high. – Glorfindel Apr 28 '17 at 8:17
• @Glorfindel so what is right way to go? – user373141 Apr 28 '17 at 8:21
• Sorry, too busy right now to write a detailed answer. I'm sure somebody else will do it. – Glorfindel Apr 28 '17 at 8:22