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05 Nov 2007, 10:10
Kudos
Bookmarks
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from other two?
(A) 24
(B) 36
(C) 72
(D) 144
(E) 216
I tried to solve this question with the help of great probability approach from KillerSquirrel (thanks for such a valuable information
) but got some problems with his explaination.
Question states not to have any ZEROs, therefore available digits are 1-9.
Total available digits for any place are 9 instead of 10.
using the probability method:
N= 1,000 - 100 = 900
P1 = 1/9 * 8/9 = 8/81
P2 = 8/9 * 1/9 = 8/81
P3 = 8/9 * 1/9 = 8/81
P = P1+ P2+ P3 = 24/81
N*P = 900 * 24/81 = 266.67
There is no answer choice matching this answer
(A) 24
(B) 36
(C) 72
(D) 144
(E) 216
I tried to solve this question with the help of great probability approach from KillerSquirrel (thanks for such a valuable information
) but got some problems with his explaination.
Question states not to have any ZEROs, therefore available digits are 1-9.
Total available digits for any place are 9 instead of 10.
using the probability method:
N= 1,000 - 100 = 900
P1 = 1/9 * 8/9 = 8/81
P2 = 8/9 * 1/9 = 8/81
P3 = 8/9 * 1/9 = 8/81
P = P1+ P2+ P3 = 24/81
N*P = 900 * 24/81 = 266.67
There is no answer choice matching this answer
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 below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
05 Nov 2007, 11:02
Kudos
Bookmarks
Is the answer E
We have 9 digits (excluding 0)
We have 3 different condition in which the requirement can be met
1. first and second digits are same and the third is different
2. Second and third are same and first is different
3. First and third are same and the second is different.
Now consider that the two digits which are same as one.
So we have 9 * 8 = 72 integers
For each condition mentioned above there are 72 integers
Therefore 72 * 3 = 216
We have 9 digits (excluding 0)
We have 3 different condition in which the requirement can be met
1. first and second digits are same and the third is different
2. Second and third are same and first is different
3. First and third are same and the second is different.
Now consider that the two digits which are same as one.
So we have 9 * 8 = 72 integers
For each condition mentioned above there are 72 integers
Therefore 72 * 3 = 216
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!
