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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”?

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

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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”?

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

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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 [#permalink]

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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

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

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

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

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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