A “descending number” is a three-digit number, such that the units dig

Question

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

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10

broall · Accepted Answer

From 100 to 999, there are 999 - 100 + 1 = 900 three-digit numbers,

From 10 unit digits 0, 1, 2, ..., 9, select 3 units digits. There are totally $10C3=\frac{10!}{3!7!}=\frac{8*9*10}{2*3}=4*3*10=120$

From each selection of 3 units digits, we could create a “descending number”. Hence, there are totally 120 “descending number”.

The probability is: $\frac{120}{900}=\frac{2}{15}$

Answer C

Sahilpandey · Answer

In selection of 10digits, where we can choose 3 digits. Also remember in set of 3digits there is only one possible set which can be in ascending or descending order. Example for a set {1,2,3} we have 123, 132, 213, 231, 312, 321.
So for descending we have a choice C(10,3) = 120

So 120/900 = 2/15

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@s · Answer

Hi 
My approach is -

list of numbers to make a 3 digit number in descending order - 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 (total 10 numbers).
x > y > z
so x can be filled with 8 possible number because we need 2 number y and z after that. hence, total number of ways in which a descending number can be made is 8 x 2 x 1.
and total number of ways to make 3 digit numbers are 10C3 (120)

8 x 2 x 1 / 120 = 16 / 120
2/15

fskilnik · Answer

{\rm{descending}} = \underline a \,\,\underline b \,\,\underline c \,\,\,\left\{ \matrix{
 \,a \in \left\{ {\,1,2, \ldots ,9\,} \right\} \hfill \cr 
 \,b \in \left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr 
 \,c \in \left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr 
 \,c < b < a \hfill \cr} \right.

?\,\, = \,\,P\left( {{\rm{descending}}} \right)

{\rm{total}}:\,\,\,9 \cdot 10 \cdot 10\,\,{\rm{equiprobable}}\,\,\left( {{\rm{positive}}} \right)\,\,3{\rm{ - digit}}\,{\rm{numbers}}

{\rm{favorable}}\,\,{\rm{:}}\,\,\,{\rm{C}}\left( {10,3} \right)\,\,\,\,\left( * \right)\,

? = {{C\left( {10,3} \right)} \over {9 \cdot 10 \cdot 10}} = {{10 \cdot 9 \cdot 8} \over {3!\,\, \cdot 9 \cdot 10 \cdot 10\,}} = {8 \over {6 \cdot 10}} = {2 \over {15}}

(*) Whenever you choose 3 different digits among the digits from 0 to 9, you are selecting exactly one possibility of descending number, and vice-versa.

Example 1: choose 0, 3, 7 then you get (putting them in decreasing order) the descending number 730 
Example 2: choose 1, 9, 8 then you get (putting them in decreasing order) the descending number 981

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

BrentGMATPrepNow · Answer

Let's use some  counting methods  to solve this.

P(selected number is "descending") =  total # of descending numbers / total # of 3-digit numbers

total # of 3-digit numbers   
3-digit numbers go from 100 to 999 inclusive
A nice rule says:  the number of integers from x to y inclusive equals y - x + 1 
999 - 100 + 1 =   900

total # of descending numbers  
First recognize that, in a descending number, all  3 digits are different  (due to the condition that units digit < tens digit < hundreds digit)
So, let's first choose 3 different digits (from 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9)
Since the order in which we choose the numbers does not matter, we can use COMBINATIONS. 
We can select 3 digits from 10 digits in 10C3 ways (= 120 ways)
 If anyone is interested, the video below shows you how to calculate combinations (like 10C3) in your head

IMPORTANT: At this point, we've selected 120 sets of 3 different digits. 
Since there's only 1 way to arrange those 3 digits in descending order, we can say that, for each set of 3-different digits, there is exactly ONE descending number, 
For example, let's say one of the 3-digit sets is {5, 1, 8}. There is only one way to arrange these 3 digits into a descending number (158)
Since we have 120 different 3-digit sets, there must be   120   descending numbers

So, P(selected number is "descending") =   120  /  900  
= 2/15

Answer: C

RELATED VIDEO FROM OUR COURSE
 https://www.youtube.com/watch?v=05a0A7vjG8I

ScottTargetTestPrep · Answer

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

learnwithrohan · Answer

we have to use three different digits to get the favorable number.
let abc be the digit such that a>b>c

now , there are tow options 
Case 1: none of the digits is 0
in this case , we simply select 3 different digits from 9 digits. once selected , we will deliberately assign the highest value to a , followed by b and c , thereby maintaining the question condition 
9c3 = 84

Case 2: one of the digits is 0 
now only the units digit can be 0 as the order of numbers is abc where a>b>c

so number becomes ab0
again to select a and b we have 9c2 = 36 options

so total : 120
probability : 120/900 => 2/15

Rohan
Private Tutor  Rohan'S  Academy

Adit_ · Answer

All in descending order means 1 way out of all arrangement.
Effectively it means:
Choose 3 digits from 10 possible digits.
10C3/900 = 120/900 =  2/15 .

Answer:  Option C

Adit_ · Answer

There is no need to divide it into 2 cases as we are considering only choosing the 3 digits and there is no arrangement involved.
Thus 10C3 directly works.

A “descending number” is a three-digit number, such that the units dig

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