varunmaheshwari wrote:
fortunetellerz,
Thanks for the explanation but this is almost the same way i solved it. Is there any more simpler and easier way to solve it?
I would suggest to split the possible alternatives of 7-digit even numbers less than 3,000,000 and calculate the arrangements. I like to use the Anagram Grids (you can review this method
Manhattan GMAT guide 5. Number Properties, Chapter 3. Combinatorics).
First:
1,XXX,XX2 - remaining numbers: 2,3,5,5,6 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)
Second:
1,XXX,XX6 - remaining numbers: 2,2,3,5,5 (Five (numerator) numbers which include two repeated digits (2,2 and 5,5) (denominator): 5!/(2!*2!) = 30)
Third:
2,XXX,XX2 - remaining numbers: 1,3,5,5,6 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)
Fourth:
2,XXX,XX6 - remaining numbers: 1,2,3,5,5 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)
Then adding 60+30+60+60=210