Here's maybe a simpler way of looking at the problem (than aa_b's correct solution):
First, how many such numbers are there between 000000 and 999999 inclusive? It's just:
(9+6)!/(9!6!) = 5005
One way to see this is to represent each qualifying number pictorially by a sequence of dots (6 in total) and bars (9 in total). There's a bijection between the set of all qualifying numbers, and the set of all such permutations of 15=6+9 symbols, since the bars divide the dots into 10 compartments, with the number of dots in each compartment giving the number of 0's, 1's,...,9's respectively. For example:
......||||||||| --> 000000
|||||||||...... --> 999999
etc.
To use this result for the actual set, 000000 (which satisfies the criterion) is replaced by 1000000 (which doesn't), so the desired total is 5005-1 = 5004, giving the probability:
5004/1000000 = 1251/250000
2006-07-09 12:30:12
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answer #1
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answered by shimrod 4
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let me try, But this may be the most ugly solution:
1-10 all nine single digital integer are ok. (1+1+1...=9)
11-100 there are: (9+8+7+...1 = 9*(9+1)/2=45)
11,12...19 (9)
22,23,...29 (8)
33,34,...39 (7)
...
99 (1)
101-1000 there are 45+36+28+21+15+10+6+3+1=165
111, 112, 113,...119 (9)
122, 123, 124,...129 (8)
...
199 (1)
222, 223,224,...229 (8)
233,...........................(7)
244............................(6)
..
299 (1)
----------------------------------------------
1001-10000
(165+120+84+56+35+20+10+4+1)=495
1111-1999 (165)
2222-2999 (165-45)=120
3333-3999 (165-45-36)=84
......
9999
-------------------------------------------------------------
10,001-100,000
495+330+210+126+70+35+15+5+1
=1287
100,001-1,000,000
1287+792+462+252+126+56+21+6+1
=3003
Total number=9+45+165+495+1287+3003
=5004
2006-07-08 17:02:17
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answer #2
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answered by aa b 1
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I will outline the general idea
Consider the number is abcde
Now a cannot be 0
b >=a
c>=b
d>=c
e>=d
a can take up any of the values between 1 to 9
And the rest shall follow.
2006-07-08 13:08:28
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answer #3
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answered by ag_iitkgp 7
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is this an integer?
hehe, should have read the question a little more closely :)
I'm not really seeing an easy way to show this. Maybe I'll get back to you later.
2006-07-08 13:02:07
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answer #4
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answered by Eulercrosser 4
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Man... now I wish I'd paid more attention to math in school!
2006-07-08 13:04:13
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answer #5
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answered by Jylsamynne 5
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Is that my left, or "stage" left?
2006-07-08 13:08:03
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answer #6
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answered by oscarsnerd 2
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