English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Is it as simple as [(10)x(9)(8)(7)(6)(5)(4)(3)]? i.e. There are 10 ways to fill two digits so that they are the same. 9 ways to fill the 3rd; 8 ways to fill the 4th, and so on. This gives 1,814,400

2007-05-22 14:54:52 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

Edit for additional information: Numbers can start with "0". So, it's more like a 9-digit combination lock problem. e.g. 024278569 is a valid choice.

2007-05-22 15:15:08 · update #1

4 answers

Here's how I'd do it.
There are three combinations at work
1) choosing 8 digits out of the 10 to form the 9 digit number
2) choosing a repeat digit
3) arranging the digits

No. of ways of choosing 8 digits out of 10 = C(10,8) = 45
No. of ways of choosing a repeat digit = 8
No. of ways of arranging the 9 digits (with one repeated)
= 9! / 2

So final answer = 9! / 2 * 45 * 8
= 65,318,400

It can also be verified that no combination gives double answers ie nothing is counted twice.

2007-05-22 19:05:01 · answer #1 · answered by Dr D 7 · 1 0

It's:
((10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 8) * 8) * 9/10
= 104509440

That is, 10 for the first, 9 for the second, etc. Then 8 for the last as it can be the same as any of the previous 8. Multiply it by 8 again as the duplicate digit can go anywhere except before and after the digit it's the same as gives the same number. Multiply 9/10 because 1/10 times the first digit is 0

2007-05-22 14:58:47 · answer #2 · answered by Anonymous · 0 0

There are 10 ways to choose the repeated digit, (9 choose 2) ways to set the position of the 2 repeated digits, ((10-1) choose 7) ways to set the choices of the nonrepeated digits, and 7! ways to arrange them in the 7 spaces left in the number. This makes a total of 10*36*36*5040=65318400 combinations.

2007-05-22 15:48:36 · answer #3 · answered by Anonymous · 0 0

0.

if 8 of the digits are different, then what can the 9th digit be? if it is the same as one of the previous 8 digits, then there are only 7 different digits as 2 would be the same and 7 would be different. and if the 9th digit were different form the previous 8, then there would be 9 different digits and not 8.

the only way oyu could have a number with exactly 8 different numbers is if it were a 10 digit number instead of 9.

2007-05-22 15:12:06 · answer #4 · answered by Anonymous · 1 1

fedest.com, questions and answers