A â€œdescending numberâ€� is a three-digit number, such that

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Oct 03, 2018 12:26 am

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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

OA C

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Wed Oct 03, 2018 12:26 am

nonplus2: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Mon May 07, 2018 8:47 am; Location: India; GMAT Score:610

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by nonplus2 » Wed Oct 03, 2018 2:14 am

BTGmoderatorDC 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

OA C

Source: Magoosh

Excellent question!!

So, we have numbers of the form:
987, 986 ... 980; 976, 975, ..., 970;910
876, 875...870; .... 810
...
210

Looking at the sequence mentioned above, we can observe that the count of numbers is itself an arithmetic series for a given series.
i.e 987, 986 ... 980 - total 8 terms, 976, 975, ..., 970 - total 7 terms ..... 910 - 1 term
likewise 876, 875,....870 - total 7 terms, 865, 864, ..860 - total 6 terms
....
210 - 1 term
In other words, we have (1 + 2 + 3 + ... + 8) + (1 + 2 + ... 7 ) + 1

We can use summation to solve this
(n = 1 to 8)Î£n*(n+1)/2

1/2*{ (n = 1 to 8)Î£n^2 + (n = 1 to 8)Î£n } # using formula : n*(n+1)*(2n+1)/6 for sum of squares

1/2*{ 8*9*17/6 + 8*9/2 } = 120

Total 3 digit numbers we have are 999 - 100 + 1 = 900.

Hence, the probability = 120/900 Dividing the numerator and denominator by 60 = 2/15

Correct answer is C

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Wed Oct 03, 2018 5:14 am

BTGmoderatorDC 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

Source: Magoosh

Beautiful problem, BTGmoderatorDC!
$${\rm{descending}} = \underline a \,\,\underline b \,\,\underline c \,\,\,\left\{ \matrix{
\,a \in \left\{ {\,1,2, \ldots ,9\,} \right\} \hfill \cr
\,b \in \left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr
\,c \in \left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr
\,c < b < a \hfill \cr} \right.$$
$$?\,\, = \,\,P\left( {{\rm{descending}}} \right)$$
$${\rm{total}}:\,\,\,9 \cdot 10 \cdot 10\,\,{\rm{equiprobable}}\,\,\left( {{\rm{positive}}} \right)\,\,3{\rm{ - digit}}\,{\rm{numbers}}$$
$${\rm{favorable}}\,\,{\rm{:}}\,\,\,{\rm{C}}\left( {10,3} \right)\,\,\,\,\left( * \right)\,$$
$$? = {{C\left( {10,3} \right)} \over {9 \cdot 10 \cdot 10}} = {{10 \cdot 9 \cdot 8} \over {3!\,\, \cdot 9 \cdot 10 \cdot 10\,}} = {8 \over {6 \cdot 10}} = {2 \over {15}}$$

(*) 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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Oct 03, 2018 6:54 am

BTGmoderatorDC 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, we have a free video on calculating combinations (like 10C3) in your head: [url] https://www.gmatprepnow.com/module/gmat- ... /video/789

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Apr 07, 2019 5:29 pm

BTGmoderatorDC 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

OA C

Source: Magoosh

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 â†’ number of descending numbers = 1

hundreds digit = 3 â†’ number of descending numbers = 1 + 2 = 3

hundreds digit = 4 â†’ 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

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Founder and CEO
[email protected]

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