2006-09-26 18:44:30 · 15 answers · asked by Omar 1 in Science & Mathematics ➔ Mathematics
e.g
1234567890
2134567890
etc
2006-09-26 18:54:20 · update #1
Beacuse you want a 10 digit number 1st digit cannot be 0
so number of ways = 9
the2n 2nd digit number of way =8
3rd digit = 7, hence forth.
so number of ways = 9 * 9*8*7*6*5*4*3*2*1 = 9* 9!
ways or 362880* 9 ways
2006-09-26 19:08:22 · answer #1 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
To answer your question, certain basic assumptions are to be made as below:
1. All the ten digits are different i.e. there is no repetition of numbers in digits, since it will get into innumerable possibilities, all of which cannot be addressed here.
2. All the new numbers are of 10 digits only.
With the above assumptions the answer will be as below:
With 10 digits, the number of numbers that can be formed is 10! i.e. 10x9x8x7x6x5x4x3x2x1 = 3628800.
Now, of these, all the numbers starting with 0 will be treated as nine digit numbers. These will have to be deducted. The no. of nine digit numbers will be 9x8x7x6x5x4x3x2x1 = 362880.
Since the question is for rearranging, the the original number will have to be counted out in rearrangement.
Hence the number of ten digit numbers that can be formed i.e. number of rearrangements to form new numbers, that are possible by rearranging the ten digits, will be 3628800-362880-1 = 3265919.
This is the solution based on the assumptions stated by above by me.
2006-09-26 21:49:22 · answer #2 · answered by Manindomb 2 · 0⤊ 0⤋
Somewhere between 0 and 10! -1.
If the 10 digits are all the same, there are no ways you can rearrange the digits to get a new number. If the digits are all different, then it'd be 10! -1.
2006-09-26 18:55:38 · answer #3 · answered by Joe C 3 · 2⤊ 0⤋
if you mean that you can only change the place of digits then this question can have different answers depending on what digits are for example if all the digits are the same then you will have only one choice. and the bigest number can be 10!-1 this happens when all the digits are different except two of them that will be similar for example if the number is 1123456789 you will have 10 choice for each digit but because we have two number 1 so we will have two similare results so the answer is 10!-1
so the answer will be a numbe between 1 and 10!-1
2006-09-26 22:09:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋
an outstanding decision is divisible by ability of 9 if and on condition that the sum of the digits is divisible by ability of 9. If we could use all 4 numbers as quickly as and purely as quickly as, then there is not any answer with the aid of fact the sum of the digits might consistently be 3+4+5+6 = 18. If we are allowed to repeat digits, then the main important extensive type we are able to make is 6666, the sum of whose digits is 24. that is no longer divisible by ability of 9, for this reason the respond is 6666.
2016-10-18 01:27:54 · answer #5 · answered by ? 4 · 0⤊ 0⤋
1) if you could use same number in diffrent digit
9*10*10*10*10*10*10*10*10*10=9*10^9
2) if in each digits you have to use the number that is not use in another digits
9*9*8*7*6*4*3*2*1=9*9!
2006-09-26 19:10:25 · answer #6 · answered by paymanns 2 · 0⤊ 0⤋
Notice the exactitude of Joe's answer. You asked how many ways can we RE-ARRANGE the digits. Since there are 10! ways the digits can be arranged, there are 10! - 1 ways they can be re-arranged.
2006-09-26 18:58:41 · answer #7 · answered by ? 6 · 1⤊ 0⤋
i think the answer is 10p9. where p stands for permutation. it means that except 0 all other 9 digits can be arranged in 10 diff ways
2006-09-26 19:01:48 · answer #8 · answered by kevin 2 · 1⤊ 0⤋
100
2006-09-26 18:51:28 · answer #9 · answered by speak_your_mind 3 · 0⤊ 0⤋
depends entirely on what the numbers are but up to 100 different combos if all the numbers are used.
2006-09-26 18:47:20 · answer #10 · answered by Anonymous · 0⤊ 2⤋