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.
