Bunuel wrote:
How many integers n between 100 and 299 are there such that the hundreds digit of n is less than the tens digit of n and the tens digit of n is less than the units digit of n ?
A. 48
B. 49
C. 50
D. 51
E. 52
Hi,
We can divide the whole range into the two parts: 100 to 199 and 200 to 299.
100 to 199Hundreds digit is fixed in this case,i.e. 1. Tens digit can be 2,3,4,5,6,7, and 8.
For example, when tens digit is 2 we have the following possibilities:
Attachment:
Number_counting1.png [ 4.05 KiB | Viewed 2166 times ]
Number of ways = 1*1*7 = 7 ways.
Similarly, when tens digit is 3 we have the following possibilities:
Attachment:
Number_counting2.png [ 3.9 KiB | Viewed 2160 times ]
Number of ways = 1*1*6 = 6 ways.
Hence, the total number of ways = \(7 + 6 + \dots + 1 = \frac{7\times8}{2} = 28\) numbers.
200 to 199Hundreds number is 2. Tens digit can be 3,4,5,6,7, and 8. Hundreds digit can be 4,5,6,7,8, and 9.
Hence, the total number in this case = \(6 + 5 + \dots + 1 = \frac{6\times7}{2} = 21\) numbers.
Total number = 28 + 21 = 49 numbers.
Thanks.