avellanjc
Shouldnt the not 2 cases be:
5*9*9= 405
The word between doesnt include endpoints. So for the hundreds digits we have (3,4,5,6,7), For the the tens we have 10 options minus the 2 option. So 9 options for the tens. And for the units same as tens.
Please clarify.Thanls in advance.
Juan C.Avellan
Archit3110
How many whole numbers between 100 and 800 contain the digit 2?
A. 200
B. 214
C. 220
D. 240
E. 248
Total digits 700
Not with 2
6*9*9 :486
With 2 700-486; 214
Option B
Posted from my mobile device
Total numbers excluding both 100 and 800 = 699.
Now, the count of numbers without the digit 2:
6 choices for the hundreds digit (1, 3, 4, 5, 6, 7),
9 choices each for the tens and units digits (0, 1, 3, 4, 5, 6, 7, 8, 9),
So total = 6 * 9 * 9 = 486.
However, this count includes 100 (since we can have 1 as the hundreds digit and 0 for both the tens and units digits), which is not allowed. However, it does not include 800 (since 8 is not one of the allowed choices for the hundreds digit), which is already outside the range. So we subtract 1 to remove 100: 486 - 1 = 485.
Therefore, the number of integers with the digit 2 = 699 - 485 = 214.
In the original solution, Archit3110 overcounted the total as 700 (1 too many) and also overcounted the "without 2" cases as 486 (again, 1 too many), but the two errors canceled each other out, so the final answer of 214 was still correct.