1
$\begingroup$

How many 4 digit numbers greater than or equal to 3000 and less than 8000 can be formed with no repetition in their digits?

$\endgroup$

closed as off-topic by Watson, JMP, Claude Leibovici, iadvd, Joey Zou Aug 24 '16 at 16:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, JMP, Claude Leibovici, iadvd, Joey Zou
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

It's: 5 * 9 * 8 * 7

  1. You can choose 5 numbers on the first place (3, 4, 5, 6, 7)
  2. You can choose 10 - 1 (choosen in 1.) numbers -> 9
  3. 9 - 2 (choosen in 1 and 2) -> 8
  4. 9 - 3 (3 numbers on previous positions) -> 7
$\endgroup$
  • $\begingroup$ Oh okay that deffinately helped me thanks, I believe actually it would be slot one is 5 two is 9 (digits 0 through 9 excluding the first digit) then slot three is 8 and slot seven would be 7? Leaving me with an answer of 2520? $\endgroup$ – Doug Mccoppen Aug 24 '16 at 10:14
  • $\begingroup$ @DougMccoppen you're right. I've edited. $\endgroup$ – bawq Aug 24 '16 at 10:21
  • $\begingroup$ Thank you so much for the help :) $\endgroup$ – Doug Mccoppen Aug 24 '16 at 10:21
0
$\begingroup$

For all numbers between $3000$ and $3999$, you first choose the first leading "3" then it remains three "slots" which cannot be filled by a "3", this leaves you $\frac{9!}{6!}$ possibilities. (Since you have 9 numbers to choose without repetition and the order matters)

Same argument for $[4000,4999], ... , [7000,7999]$

$\endgroup$
  • 2
    $\begingroup$ Isn't it 9!/6! possibilities? $\endgroup$ – bawq Aug 24 '16 at 10:20
  • $\begingroup$ My bad, I always forget there are 10 digits :p $\endgroup$ – Zubzub Aug 24 '16 at 11:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.