a maths question( seriously)?

Question

in the integer 3,589 the digits are all different and increase from left toright. how many integers between 4, 000 and 5,000 have digits that are all different and that increase from left to right?
I really don't feel like listing all the numbers and count, so anyone has a simple and explainable method for this one?
thx a lot

actually, i am not fooling around, it's just a maths question which i really want to solve, so, you guys wanna help?
please tell me the whatever forumula, thx?

Answer 1

It's not hard.
4567 4568 4569
4578 4579
4589
4678 4679
4689
4789

That's all. There are ten numbers.

Answer 2

The method is "fundamental principle of counting" in Permutations and Combinations

There are 1 X 9 X 8 X 7 = 504 numbers between 4000 n 5000 which are having all digits different

The method is simple enough there is only "1 way" to fill thousands place, since its definitely starting with the digit 4. Excluding 4, there are 9 digits, and there are "9 ways" to fill the hundreds position. so out of the 10 digits 1234567890, 2 are occupied. remaining 8 digits are available for filling tens place, so there are "8 ways" to fill up the tens place. Again, now out of total ten digits, 3 are occupied, so there are 7 digits left to fill units place, so there are "7 ways" of filling units place


"1 way" X "9 ways" X " 8 ways" X "7ways" = 504 ways, so there are total 504 total combination beginning with 4 and numbers different

Answer 3

You need to look at each digit in the integer to see how many candidate numbers there are. The first digit has to be 4, so there is only one option. The second digit cannot be greater than 7, otherwise numbers have to repeat (4899) which gives us three numbers (5,6,7). Similarly the third digit cannot be higher than 8, (6,7,8) and the final number can be 7,8,or 9. Thus we get 1+3+3+3= 10 numbers.

Answer 4

your thousands digit is four so you have digits 5 or 7 as the hundreds.

a. for 5 hundreds, you could only have 6 to 8 as tens
b. for 6 hundreds, you could only have 7 or 8 as tens

to summarize it

3 * 6 * 6 = 108

assuming that the highest digit is 9 not zero

:) goodluck

may i ask bartimaeus why would he start with 9? since only digits 5-9 is greater than 4 but you can't use 8 & 9 since it would be reserved for the tens and ones digit

but she said between 4000 - 5000 doesnt that mean that the thousands digit be equal to 4. but i think someone up there already answered the question right

:)

Answer 5

for the first digit you can only use 4.
for the 2nd digit, you have 5 6 7

for the 2nd digit :
if 5 : for the 3rd & 4th digits you have 67 68 69 78 79 89
if 6 : for the 3rd & 4th digits you have 78 79
if 7 : for the 3rd & 4th digits you have only 89

it makes 9 numbers....

Answer 6

I'm not sure about a formula of hand.
But start with a 4 an increasing the numbers this is what I've come up with, ten numbers in all.

4567, 4568, 4569, 4578, 4579, 4589
4678, 4679, 4689
4789

Answer 7

Yeah, there is a formula for how many ways n things may be chosen in order from an ordered list of p things.

But I don't remember it off the top of my head and I'm too damned tired to go looking for it.

G'night All. I'm gettin' horizontal ☺


Doug

Answer 8

The first integer should be 4 which is not here. therefore is no integer which is between 4,000 and 5,000

Answer 9

4c4 combinations............24 to be exact....now you think about listing these numbers. I just had a haircut. don wanna look like einstein listing them out.

Answer 10

nice try dude...people aint that foolish here!!