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24 Aug 2003, 14:36
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Hello akamai
i am referring to the following problem.
How many three digit numbers are there so that they can have 2 equal digits and the other digit different from these two?
paraphrasing your explanation..
10 pairs of numbers combine with 9 other numbers...total numbers ...
10*9 = 90
next there are three different ways in which the number can be arranged...AAB, ABA, BAA...so the total is 270
Next, we have to remove numbers that start with 0.
here you mention, since this is a "symmetric problem", exactly 1/10 of the numbers start with 0. so the final answer is 270 *0.9 =243
I need some help with the "symmetric " part ....can u please explain in brief.
Thanks
Praetorian
[/b]
i am referring to the following problem.
How many three digit numbers are there so that they can have 2 equal digits and the other digit different from these two?
paraphrasing your explanation..
10 pairs of numbers combine with 9 other numbers...total numbers ...
10*9 = 90
next there are three different ways in which the number can be arranged...AAB, ABA, BAA...so the total is 270
Next, we have to remove numbers that start with 0.
here you mention, since this is a "symmetric problem", exactly 1/10 of the numbers start with 0. so the final answer is 270 *0.9 =243
I need some help with the "symmetric " part ....can u please explain in brief.
Thanks
Praetorian
[/b]
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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24 Aug 2003, 16:42
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praetorian123
The symmetric part means every number is equally likely to be in the first digit spot. Since there are 10 numbers, the likelyhood of a zero being in the first spot is simply 1/10.
25 Aug 2003, 07:11
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In this problem, each number 100 - 999 can fall into one of three categories.
(1) All three digits are the same
(2) All three digits are different
(3) Two digits are the same and one is different
One method is to calculate the total # of numbers, and subtract (1) and (2) to leave (3)
Total # of numbers is 900 (999 - 100 + 1)
(1) there are 9 (111, 222, etc...)
(2) 9 possibilities for the first (can't be zero), 9 possibilities for the second (can't be the first), 8 possibilities for the last (can't be either of the first two). 9*9*8 = 648
900 - 9 - 648 = 243
(1) All three digits are the same
(2) All three digits are different
(3) Two digits are the same and one is different
One method is to calculate the total # of numbers, and subtract (1) and (2) to leave (3)
Total # of numbers is 900 (999 - 100 + 1)
(1) there are 9 (111, 222, etc...)
(2) 9 possibilities for the first (can't be zero), 9 possibilities for the second (can't be the first), 8 possibilities for the last (can't be either of the first two). 9*9*8 = 648
900 - 9 - 648 = 243
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
