Bunuel wrote:
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?
A. 729
B. 243
C. 108
D. 88
E. 77
Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for
XYZ, where
Z is an even number.
If the first digit (X) is 7 or 9 (
2 values) then the third digit (Z) can take all
4 values (2, 4, 6, or 8) and the second digit (Y) can take
7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;
If the first digit (X) is 8 (
1 value) then the third digit (Z) can take only
3 values (2, 4, or 6,) and the second digit (Y) can take
7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;
Total: 56+21=77 numbers.
Answer: E.
Hope it's clear.
Hi Bunuel,
If the first digit (X) is 7 or 9 (
2 values) then the third digit (Z) can take all
4 values (2, 4, 6, or 8) and the second digit (Y) can take
7 values (9 minus two digits we already used).
Here what about 2 as second digit. For eg: 722. In this case second digit can't take 2. How are these numbers (such as 744,766,788,922,944) removed from 2*4*7 = 56 numbers?
If the first digit (X) is 8 (
1 value) then the third digit (Z) can take only
3 values (2, 4, or 6,) and the second digit (Y) can take
7 values (9 minus two digits we already used).
Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?