Bunuel wrote:
A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
We see that the hundreds digit can be any digit from 2 to 9.
If we start with the hundreds digits as 2, then the tens digit must be 1, and the units digit must be 0. We have 1 descending number when the hundreds digit is 2.
If the hundreds digits is 3, then if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 3 descending numbers when the hundreds digit is 3.
Now, if the hundreds digit is 4, then if the tens digit is 3, the units digit can be 2, 1, or 0; if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 6 descending numbers when the hundreds digit is 4.
From this point, we can see a pattern:
hundreds digit = 2 → number of descending numbers = 1
hundreds digit = 3 → number of descending numbers = 1 + 2 = 3
hundreds digit = 4 → number of descending numbers = 1 + 2 + 3 = 6
So the number of descending numbers when the hundreds digit is 5 is 10, 6 is 15, 7 is 21, 8 is 28, and 9 is 36. Therefore, there are a total of 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 three-digit descending numbers.
Since there are 999 - 100 + 1 = 900 three-digit numbers, the probability of picking a descending number is 120/900 = 12/90 = 2/15.
Answer: C