Question #51dfa

1 Answer
Nov 30, 2017

#4200# distinct values (I have assumed that the initial digit could be a zero).

Explanation:

There are #color(white)("x")_10C_3# ways of selecting which #3# digits (out of a possibility of #10# digits) will make up the set #{a,b,c}#. Notice that once this set has been selected, the rule that #a < b < c# has only one possibility for the sequence #abc#.

Once #{a,b,c}# has been selected there are #color(white)("x")_7C_3# ways of selecting which #3# digits (out of the remaining #7# digits) will make up the set #{d,e,f}# and, again, once these digits have been selected the rule that #d > e > f# dictates only one possibility for the sequence #def#

Combining these observations, we have:
Number of distinct values
#color(white)("XXX")=color(white)(".")_10C_3 xx color(white)(".")_7C_3#

#color(white)("XXX")=(10!)/(3! * 7!) xx (7!)/(3! * 4!)#

#color(white)("XXX")=(10 * 9 * 8)/(3 * 2 * 1) xx(7 * 6 * 5)/(3 * 2)#

#color(white)("XXX")=120 xx 35 (=60xx70)#

#color(white)("XXX")=4200#