How many even 3 digit integers greater than 700 with

NandishSS 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

Case 1: Hundreds digit is 7 or 9
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (2, 4, 6, or 8)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
2*4*7 = 56.

Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 3. (2, 4, or 6)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
1*3*7 = 21.

Total ways = Case 1 + Case 2 = 56+21 = 77.

The correct answer is E.

GMATGuruNY wrote:

NandishSS 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

Case 1: Hundreds digit is 7 or 9
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (2, 4, 6, or 8)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
2*4*7 = 56.

Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 3. (2, 4, or 6)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
1*3*7 = 21.

Total ways = Case 1 + Case 2 = 56+21 = 77.

The correct answer is E.

Hi Mitch,

I tried to solve it as follows:
Number of options for the hundreds digit = 3. (7 or 8 or 9)
Number of options for the units digit = 3. (2, 4, or 6) - excluding 8
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply: 3*3*7= 63

Where did I go wrong?

Thanks

Mo2men wrote:Hi Mitch,

I tried to solve it as follows:
Number of options for the hundreds digit = 3. (7 or 8 or 9)
Number of options for the units digit = 3. (2, 4, or 6) - excluding 8
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply: 3*3*7= 63

Where did I go wrong?

Thanks

The red portion miscounts the number of options for the units digit.
The red portion is valid only if the hundreds digit is 8, in which case the units digit can be 2, 4, or 6, yielding a total of 3 options.
But if the hundreds digit is 7 or 9, then the units digit can be 2, 4, 6, or 8, yielding a total of 4 options.
Thus, it is incorrect to say that -- in every case -- the number of options for the units digit = 3.