Permutations\Combinations from digits

Using digits 1,2,3,4,5,7 only, how many numbers could be made that are between 2500 and 5000, if a digit is not repeated?

(If it had been 2000-5000, that would have been easy!!!)

The answer at my level should only use $^nP_{r}$, $^nC_{r}$ and\or factorial notation.

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HINT: how many are between 3000 and 5000 (inclusive)? For 2500 and 3000, you are really asking how many are between 500 and 999 (inclusive), not using the digit 2 either.

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3000-5000: 120 numbers ------- 500-999: 24 numbers------ total 144 correct! –  schooler Aug 22 '11 at 13:28

Since you say 2000 to 5000 is easy, all you have to do is do that and then subtract the ones between 2000 and 2500. And how many of those are there? Well, how many choices for the first digit? for the second digit? the third? the fourth? and what do you do with those numbers?

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we need to find numbers between $2500$ to $5000$ with given digits $1,2,3,4,5$ and $7$ without repeating.
so, first let us think about the numbers started with 2 that means second digit must be 5 or 7 and remain with two spaces so for them we can have 4 chances to select third digit and fourth digit has 3 chances.
so number of numbers started with 2 and satisfying conditions are product of $2,4,5=24$
like this the numbers started with 3 or 4 are product of $2,5,4,3=120$
therefore total numbers is $120+24=144$

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