Q. Out of $21$ tickets, numbered $1$ to $21$, three are selected at random find the probability that the numbers are in $AP$.
Attempt-
$$ \frac{(a-d)!}{((a-d)-1)!} \times a \times \frac{(a+d)!}{[(a+d)-1]!} $$
$$ P(E)= \frac{a(a^{2}-d^{2}) }{1330} $$
Should the answer always depend on the value taken? Or is there some other way to calculate the probability which will give pure numerical value?
You have an idea of how the solution might proceed, but what you are missing is the enumeration over all possible values of the central number.
Let's count the number of admissible triplets. Suppose we select three numbers that form an arithmetic progression from the set $\{1, 2, \ldots, 21\}$. Then, as you have observed, this triplet is expressible in the form $(a-d, a, a+d)$ for some integers $a, d$ such that $$1 \le a-d < a < a+d \le 21.$$ It should be clear that the smallest such $a$ is $2$ and the largest is $20$. For some general $a$ in this range, a bit of thought shows that there are $\min(21-a, a-1)$ choices for $d$: for example, if $a = 4$, we could choose $d \in \{1, 2, 3\}$. If $a = 15$, then you could choose $d \in \{1, 2, 3, 4, 5, 6\}$. So the number of triplets in AP is given by $$n = \sum_{a=2}^{20} \min(21-a, a-1).$$ Well, $21-a = a-1$ when $a = 11$, the midpoint of $\{1, 2, \ldots, 21\}$. So we split the sum at this value: $$n = \sum_{a=2}^{11} a-1 + \sum_{a=12}^{20} {21-a} = \frac{10(11)}{2} + \frac{9(10)}{2} = 100.$$ Finally, to get the probability, we simply divide by the number of possible triplets, which is $\binom{21}{3} = 1330$, hence the probability is $$\frac{10}{133}.$$
How would you generalize this to the case where you select from the set $\{1, 2, \ldots, m\}$?
Firstly, taking $AP$ with $d=1$
Cases are $[1,2,3], [2,3,4], ... , [19,20,21]$, ie, total cases = $19$
$AP$ with $d=2$
Cases are $[1,3,5], ... , [17,19,21]$, ie, total cases = $17$
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$AP$ with $d=10$
Cases are $[1,11,21]$, ie, total cases = $1$
So, total number of cases in which AP is formed $= 19 + 17+ ... 1 = (10)^2=100$
Total number of cases = $21\choose{3}$
Required probability = $\frac{100}{ 21\choose{3}}=\frac{100}{1330}=\frac{10}{133}$
We can choose any three out of $21$ numbers in $$\binom {21}{3} = 1330$$ ways. For the numbers to be in AP, we choose the common difference of the progression and then check the number of possible cases.
If the difference is $1$, the possible terms can belong to $\{(1,2,3) ; (2,3,4) ; \cdots (19,20,21)\} $, a total of $19$ values. If the difference is $2$, the possible terms are $\{(1,3,5) ; (2,4,6) \cdots (17,19,21)\} $. And so on, till, if the common difference is $10$, the possible terms are $\{(1,11,21)\} $.
So, total no. of group of $3$ numbers drawn from the $21$ which are in AP are: $$ 19 + 17 +\cdots 3 +1 =10^2 =100$$
Thus, we have a probability of: $$\frac {100}{1330} = \frac {10}{133} $$
Hope it helps.