Question 1082065
(i) There are {{{(matrix(2,1,7,3))=7!/3!=7*6*5}}} sequences
of {{{3}}} characters that can be made with the {{{7}}} given digits,
without repeated characters.
However, {{{1/7}}} of those sequences would have {{{0}}} as its first character,
so only the other {{{1-1/7=6/7}}} of them are three-digit numbers.
That is  {{{7*6*5*(6/7)=6^2*5=highlight(180)}}} .
Calculating it a different way,
of the {{{7*6*5=210}}} 3-character sequences,
{{{210/7=30}}} start with {{{0}}} ,
so only {{{210-30=highlight(180)}}} are three-digit numbers.
 
(ii) It is easier to start from the last digit.
For an odd three-digit number, it is necessary and also sufficient to have an odd last digit.
In the {{{7}}} digits available,
there are {{{3}}} odd digit choices for a last digit
that would make the three-digit number odd.
That leaves {{{7-1-1=5}}} non-zero digits left to use for a first digit.
Having used {{{2}}} non-zero digits for first and last digit,
you have {{{7-2=5}}} unused digits to use for a middle digit.
That gives you {{{3*5*5=highlight(75)}}} three-digit numbers,
without repeated digits, that can be made from the given digits,
and are odd.
 
(iii) There are {{{(matrix(2,1,7,3))=7!/3!=7*6*5=210}}} sequences
of {{{3}}} characters that can be made with the {{{7}}} given digits,
without repeated characters.
The ones that are three-digit numbers less than 300 start with {{{1}}} or {{{2}}} .
Only {{{2/7}}} of the 3-character sequences would have {{{1}}} or {{{2}}} as the first character,
so {{{7*6*5(2/7)=2*6*5=60}}} are three-digit numbers less than 300.
That means that the probability of forming a three-digit number less than 300 is
{{{60/180=highlight(1/3)}}} .