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CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

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?
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

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]
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 52 [0], given: 0

praetorian123 wrote:
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?
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

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]

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.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 4 [0], given: 0

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
_________________

Sept 3rd

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