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How many numbers between $200$ and $1200$ can be formed with the digits $0,1,2,3 $ (repetition of digits not allowed ) ?

$a.)\ 6\\ b.)\ 8\\ c.)\ 16\\ \color{green}{d.)\ 14}$

I divided it in $3$ digit and $4$ digit numbers.

$3$ digit numbers : $200-999$

For these their are $2\times 3\times 2=12$ ways.

$4$ digit numbers : $1000-1200$

For the first digit $\{1\}$ only $1$ way.

For the second digit $\{0,2\}$ only $2$ way

For the third digit $\{0,2,3\}$ only $3-1$ way

For the fourth digit $\{0,2,3\}$ only $3-2$ way

=$2\times 2=4$ ways.

Total=$16$ ways but the book is showing option $c.) 14$ ways

I look for a short and simple way.

I have studied maths upto $12$th grade.

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    $\begingroup$ I don't think 2 can be the second digit of the 4 digit answer. 1203 or 1230 will not qualify. $\endgroup$ Sep 3, 2015 at 1:56
  • $\begingroup$ @turkey: I think u r right $\endgroup$
    – R K
    Sep 3, 2015 at 2:05
  • $\begingroup$ is 230 allowed (because it is built by 2,3 and 0) or is it not allowed ( because it is viewed as 0230 and therefore the digit 0 is used two times)? $\endgroup$
    – miracle173
    Sep 3, 2015 at 2:30

1 Answer 1

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You can't have $2$ for the second digit in your $4$ digit number. The highest number you could form, provided you had the digits required is 1199. Any $4$ digits number with 2 for second digit is more than that upper limit.

So, your choice(in 4 digit number) for:
first digit: $1$ (you can have $1$ only)
second digit: $1$ (you can have $0$ only)
third digit: $2$ (you can have either $2$ or $3$)
fourth digit: $1$ (the one that remains)

That is: $1.1.2.1 = 2$ numbers you can make.

So, you can make $12+2 =14$ numbers in total given the condition.

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