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A “descending number” is a three-digit number, such that the units dig
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15 Aug 2017, 23:33

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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”?

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16 Aug 2017, 00:59

[quote="Bunuel"]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”?

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16 Aug 2017, 01:10

dave13 wrote:

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 “descending number” is a three-digit number, such that the units dig
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16 Aug 2017, 01:24

Sahilpandey wrote:

dave13 wrote:

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”?

you are so right, thanks! but I guess there must be a short cut to consider all such numbers, because its really time consuming... and I can`t imagine how it is doable in two minutes... there is a high risk that one number will be missing

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16 Aug 2017, 01:28

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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

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: \(\frac{120}{900}=\frac{2}{15}\)

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16 Aug 2017, 01:36

broall wrote:

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

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

hello ! I knew it , there was a shortcut method could you please break down it in more details

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

I have basic knowledge of combinations and permutations, but I cant understand how did you apply it here

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16 Aug 2017, 01:38

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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

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08 Jun 2018, 11:20

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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

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)

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03 Oct 2018, 06:20

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

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

(*) 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.
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Re: A “descending number” is a three-digit number, such that the units dig
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03 Oct 2018, 07:58

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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

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

A “descending number” is a three-digit number, such that the units dig
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03 Oct 2018, 14:37

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

in 200s-1 (210) in 300s-3 (310,320,321) in 400s-6 (410,420,421,430,431,432 1,3,6,10,15,21...is a triangular sequence for given term n, sum=n(n+1)(n+2)/6 for 8 blocks (200s-900s), n=8 and sum=120 total descending numbers 120/900=2/15 C

Re: A “descending number” is a three-digit number, such that the units dig
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16 Oct 2018, 21:45

Total Outcomes= Three-digit numbers ---> 999-100+1 = 900.

For any arrangement of three numbers, there is only one in which the numbers are arranged so that the units are less than the tens, which are less than the hundreds: 10x9x8 = 720 are the total different combinations of three different integers. Only one of the six possible arrangements (3! = 6) matches up with the restraints in the problem. Thus, 720/6 = 120.
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Re: A “descending number” is a three-digit number, such that the units dig
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07 Apr 2019, 18:32

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

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.