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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?

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  • $\begingroup$ It depends on whether 0112 is a valid 4 digit number. $\endgroup$ – Glorfindel Apr 28 '17 at 8:08
  • $\begingroup$ @Glorfindel absolutely invalid ; but how can i remove such cases? $\endgroup$ – user373141 Apr 28 '17 at 8:10
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    $\begingroup$ 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. $\endgroup$ – Glorfindel Apr 28 '17 at 8:17
  • $\begingroup$ @Glorfindel so what is right way to go? $\endgroup$ – user373141 Apr 28 '17 at 8:21
  • $\begingroup$ Sorry, too busy right now to write a detailed answer. I'm sure somebody else will do it. $\endgroup$ – Glorfindel Apr 28 '17 at 8:22
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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

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    $\begingroup$ Also this solution does not count numbers that have the digit "1" once three times and four times. $\endgroup$ – dvd280 Apr 28 '17 at 8:42
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    $\begingroup$ Elegant Solution, thank you very much $\endgroup$ – user373141 Apr 28 '17 at 9:02

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