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

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

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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”?

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10
[Reveal] Spoiler: OA

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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 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”?

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10

hello Bunuel how are you :) hope you are well!:)

you know I tried to solve your questions but arrived at totally different answer :)

the descending numbers are as follow:

210, 321, 432, 543, 654, 765, 876, 987, 320, 430, 540, 650, 760, 870, 980 total 15 numbers

the total number of three digit numbers is 999-100+1= 900

the probability that descending number is picked at random is 15/900 = reducing it I get 1/60

can you please say where I am wrong ? thank you !

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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 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) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10

hello Bunuel how are you :) hope you are well!:)

you know I tried to solve your questions but arrived at totally different answer :)

the descending numbers are as follow:

210, 321, 432, 543, 654, 765, 876, 987, 320, 430, 540, 650, 760, 870, 980 total 15 numbers

the total number of three digit numbers is 999-100+1= 900

the probability that descending number is picked at random is 15/900 = reducing it I get 1/60

can you please say where I am wrong ? thank you !
you forget to consider the nos like 310 , 431,421,430,420,410,530,530,540 etc and many more that's why you came up with a wrong answer


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

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New post 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”?

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10

hello Bunuel how are you :) hope you are well!:)

you know I tried to solve your questions but arrived at totally different answer :)

the descending numbers are as follow:

210, 321, 432, 543, 654, 765, 876, 987, 320, 430, 540, 650, 760, 870, 980 total 15 numbers

the total number of three digit numbers is 999-100+1= 900

the probability that descending number is picked at random is 15/900 = reducing it I get 1/60

can you please say where I am wrong ? thank you !
you forget to consider the nos like 310 , 431,421,430,420,410,530,530,540 etc and many more that's why you came up with a wrong answer


Sent from my iPhone using GMAT Club Forum mobile app



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 :-D

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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 16 Aug 2017, 01:28
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
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Kudos [?]: 1093 [0], given: 62

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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 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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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 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

So 120/900 = 2/15


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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 16 Aug 2017, 01:39
dave13 wrote:
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


Cant get your ideas. Could you specify more?
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Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 16 Aug 2017, 01:41
I am sending you a direct message of the explanation


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Re: A “descending number” is a three-digit number, such that the units dig [#permalink]

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New post 17 Aug 2017, 06:02
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Please refer to the attached image for detailed solution .

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Re: A “descending number” is a three-digit number, such that the units dig   [#permalink] 17 Aug 2017, 06:02
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