Kellogg MMM ThreadMaster
Joined: 29 Mar 2012
Posts: 324
Location: India
GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38

Re: Tricky question [#permalink]
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20 Jun 2012, 00:11
Hi,
1) You have to find the even (3 digit) numbers greater than 700, with nonzero digits as well as distict digits. So, available digits would be (1 to 9) Even numbers starting with 7 = 1*7*4 = 28 (Hundredth digit is 7  so, only 1 choice, unit digit can be (2, 4, 6, 8). Now tens digit will not be 7 & a digit chosen at units place  7 possibilities) Even numbers starting with 8 = 1*7*3 = 21 Even numbers starting with 9 = 1*7*4 = 28 Total numbers = 77
2) You have to find the odd (3digit) numbers greater than 800, all distict digits. Available digits would be (0 to 9) Odd numbers starting with 8 = 1*8*5 = 40 (Hundredth digit is 8  so, only 1 choice. Unit digit can be (1, 3, 5, 7, 9). Now tens digit will not be 8 & a digit chosen at units place  8 possibilities) Odd numbers starting with 9 = 1*8*4 = 40 (Hundredth digit is 9  so, only 1 choice. Unit digit can be (1, 3, 5, 7). Now tens digit will not be 9 & a digit chosen at units place  8 possibilities) Total numbers = 32
Let me know, if you need any further help on this.
Regards,
