If the number of triplet $(l,m,n)$ satisfying $1\leq l\leq m\leq n \leq 20$ is $p \choose q$ then least value of $p+q$ is? 

If the number of triplet $(l,m,n)$ satisfying $1\leq l\leq m\leq n \leq 20$ is $p
\choose q$ then least value of $p+q$ is?


My Approach:
Case $1:$ $1\leq l=m=n \leq 20$
so there are such $20$ nuumber using $20 \choose 1$
Case $2:$ $1 \leq l=m<n\leq20$
so there are $20 \choose 2$=$190$ ways
Case $3:$ $1 \leq l<m =n\leq20$
so there are $20 \choose 2$=$190$ ways
Case $4:$ $1 \leq l<m<n\leq20$
so there are $20 \choose 3$=$1140$ ways
So total Number of such order Triplets is $1540$
That is $22 \choose 3$
please let me know if i made any mistake in calculating triplets
 A: In general, for
$$ 1 \leq x_1 \leq x_2 \leq \cdot \cdot \cdot \leq x_k \leq n$$
Using stars and bars,  the number of k-tuples satisfying the above relation can be obtained as
$$\sum_{r=0}^{k-1} {n\choose k-r} \cdot {k-1 \choose r} = {n+k-1 \choose k}$$
where $k-r$ denotes the no. of distinct entities among the $k$ numbers
A: Yes your counting is correct but can be simplified.
Applying stars and bars is the easiest solution but I will come to that later. Even otherwise, note that the solution contains either
$i)$ $3$ distinct numbers in ascending order OR
$ii)$ contains $2$ distinct numbers in ascending order and one of them repeating twice OR
$iii)$ contains just one number repeated thrice.
So the answer is $ \displaystyle {20 \choose 3} + 2 \cdot {20 \choose 2} + {20 \choose 1}$
Coming to stars and bars method (wiki), here is a way I would explain it. consider writing numbers $~1 - 20~$ in ascending order. Now for numbers we choose, we put a bar above it. If we pick the same number twice, we put two bars above it. If we pick the same number thrice, we put three bars above it. Considering $19$ spaces between numbers as stars, number of ways to arrange $19$ stars and $3$ bars would be -
$ \displaystyle \frac{22!}{19! \cdot 3!} ~\text { or } ~{22 \choose 3}$
A: My dear friend @Math Lover gave you very detaily answer , so there left only one solution way , it is the tricky way .
Now , it is said that $1 \leq l \leq m \leq n \leq 20$ , so what if i manipulate the equation such that $$1\leq (l+1) \leq (m+2) \leq (n+3) \leq (20+4) \rightarrow 1 <l' <m'<n'<24$$
Now , the possible set for selecting $l'$,$m'$ and $n'$ is from $\{2,3,4,..,21,22,23\}$ . There are $22$ elements to choose. Then, select $3$ of them form this set by $C(22,3)$.They will automatically place their position because of the bijection
A: Consider instead the triplets where $1\leq l < m+1 < n+2 \leq 22$
which has a one-to-one correspondence with $1\leq l \leq m \leq n \leq 20$
The answer then immediately comes as $\binom {22}3$
