I was solving a question paper and i stuck due to some missing concepts. please help me out.I want the shortcut to solve this type of question too.
Question:Find the number of all four digit numbers whichare greater than 2367 and in which the digit are arranged in ascending order.
Split the set into two types: those numbers above 3000 and those below 3000.
The first type is easy: it cannot make use of the digits 0,1, and 2. So we pick 4 digits between 3 and 9 (and arrange them in increasing order). So this is the same as the binomial coefficient, ${7\choose 4}= 7!/(4!)(3!)=35$.
To count the second type, further split into those less than 2400 and those between 2400 and 3000. Those less than 2400 are simply seen to be $2368; 2369, 2378, 2379 $ and $2389$ giving 5 numbers.
So we are left with numbers starting with 2 and avoiding 3.
$24xx \to {5\choose2}=10$
$25xx\to {4\choose2}= 6$
$26xx\to {3\choose 2}=3$
$27xx\to {2\choose2}=1$.
Adding up we have totally $35+5+10+6+3+1$, that is 60 numbers.
The answer to your question depends on what you mean by ascending order.
The digits must be distinct:
A four digit number consisting of distinct digits written in ascending order can be formed by selecting four numbers from the sequence $123456789$. For instance, the number $2367$ corresponds to the choice $1\color{blue}{23}45\color{blue}{67}89$.
We consider cases:
The number is at least $3000$. We can choose a four digit number with distinct digits that appear in ascending order by selecting four digits from the sequence $3456789$. For instance, the choice $3\color{blue}{4}5\color{blue}{67}8\color{blue}{9}$ yields the number $4679$. The number of ways we can make such a choice is $\binom{7}{4}$ since we are choosing four of the seven numbers in the sequence.
The number is at least $2400$ but less than $3000$. To form a four digit number with distinct digits that appear in ascending order, we must select three of the six blue digits in the sequence $2\color{blue}{456789}$. For instance, the choice $2\color{blue}{4}\color{red}{5}\color{blue}{6}\color{red}{78}\color{blue}{9}$ yields the number $2578$. The number of ways of making such a choice is $\binom{6}{3}$ ways.
The number is at least $2370$ but less than $3000$. To form a four digit number with distinct digits that appear in ascending order, we must select two of the three blue digits in the sequence $23\color{blue}{789}$, which can be $\binom{3}{2}$ ways.
The number is greater than $2367$ but less than $2370$. To form a four digit number with distinct digits that appear in ascending order, we must select one of the two blue digits in the sequence $236\color{blue}{89}$, which can be done in $\binom{2}{1}$ ways, corresponding to the numbers $2368$ and $2369$.
Hence, the number of ways we can form a four digit number greater than $2367$ consisting of distinct digits that appear in ascending order is $$\binom{7}{4} + \binom{6}{3} + \binom{3}{2} + \binom{2}{1} = 35 + 20 + 3 + 2 = 60$$
The digits need not be distinct:
A four digit number whose digits appear in non-decreasing order can be represented by placing four dividers in a sequence of nine ones, so that the number of ones to the left of the $k$th divider represents the $k$th digit when reading from left to right. For instance, the number $2367$ is represented by $$1 1 \vert 1 \vert 1 1 1 \vert 1 \vert 1 1 1$$
We consider cases:
The number is at least $3000$. To select a number that is at least $3000$ whose digits appear in non-decreasing order, we must place four dividers in our list of nine ones among the six ones that appear after the first three ones. For instance, the number $3499$ is represented by
$$1 1 1 \vert \color{blue}{1 \vert 1 1 1 1 1 \vert \vert}$$
The number of such choices we can make is $\binom{6 + 4}{4} = \binom{10}{4}$ since we must choose which four places among the six ones and four dividers that appear after the first three ones will be filled by dividers.
The number is at least $2400$ but less than $3000$. To select a number that is at least $2400$ but less than $3000$ whose digits appear in non-decreasing order, we must place a divider in our list of nine ones after the first two ones, then place three more dividers among the five ones that appear after the fourth one. For instance, the choice $$1 1 \vert 1 1 \color{blue}{1 \vert 1 1 \vert \vert 1 1}$$ corresponds to the number $2577$. The number of such choices is $\binom{5 + 3}{3} = \binom{8}{3}$.
The number is at least $2370$ but less than $2400$. To select a number that is at least $2370$ but less than $2400$ whose digits appear in non-decreasing order, we must place the first divider in our list of nine ones after the first two ones, the second divider after the third one, and then place two more dividers among the two digits that appear after the seventh one. This can be done in $\binom{2 + 2}{2} = \binom{4}{2}$ ways.
The number is greater than $2367$ but less than $2370$. To select a number that is greater than $2367$ but less than $2370$ whose digits appear in non-decreasing order, we must place our first divider in our list of nine ones after the first two ones, the second divider after the third one, the third divider after the sixth one, and then place the fourth divider before or after the ninth one (corresponding to the choices $2368$ and $2369$, respectively), which can be done in $\binom{1 + 1}{1} = \binom{2}{1}$ ways.
Hence, the number of four digit numbers larger than $2367$ in which the numbers appear in non-decreasing order is $$\binom{10}{4} + \binom{8}{3} + \binom{4}{2} + \binom{2}{1} = 210 + 56 + 6 + 2 = 274$$
