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.
The symmetric part means every number is equally like
ly 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.
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