How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order? Solution for the same is given below 
Numbers starting with 12 – 7 numbers
Numbers starting with 13 – 6 numbers; 
14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1.
Thus total number of
numbers starting from 1 is given by the sum of 1 to 7 = 28.
Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21
Number of numbers starting from 3- sum of 1 to 5 = 15
Number of numbers starting from 4 – sum of 1 to 4 = 10
Number of numbers starting from 5 – sum of 1 to 3 = 6
Number of numbers starting from 6 = 1 + 2 = 3
Number of numbers starting from 7 = 1
Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers.
suggest some trick to approach without thinking about this gibberish given above?
 A: You need to choose three digits from nine.  Once the digits have been selected, the order in which they should be placed is determined.  Hence $\binom{9}{3}$ numbers are possible.
You are seeing binomial coefficients in your method too: once the first digit is fixed, there are $\binom{n}{2}$ ways to choose the remaining two digits, where $n$ is the number of digits greater than the first digit.  Equating our answers gives a binomial coefficient identity,
$$
\binom{9}{3}=\binom{8}{2}+\binom{7}{2}+\binom{6}{2}+\binom{5}{2}+\binom{4}{2}+\binom{3}{2}+\binom{2}{2},
$$
which you can think about.
A: Here is a trick:


*

*An increasing arrangement must contain unique digits

*There are $\binom{9}{3}$ combinations of $3$ out of $9$ unique digits

*Each combination has exactly $1$ increasing arrangement

*Hence the number of increasing arrangements is $\binom{9}{3}$

A: For each combination of 9 things taken 3 at a time, there is just 1 way they
can be arranged in ascending order.  So the answer is: 
9C3 or C(9,3) = 84 ways
