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

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

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

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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
• 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? – Doug Mccoppen Aug 24 '16 at 10:14
• @DougMccoppen you're right. I've edited. – bawq Aug 24 '16 at 10:21
• Thank you so much for the help :) – Doug Mccoppen Aug 24 '16 at 10:21

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 $[4000,4999], ... , [7000,7999]$

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