Question

Number of ways in which 3 tickets can be selected from a set of 500 tickets numbering from 1 to 500 so that the numbers on them are in arithmetic progression.
A. 124500
B. ${}^{500}{{C}_{3}}$
C. 62250
D. None of these

Answer 1

Hint: To solve this question, we should know the properties related to arithmetic progression and combinations. We know that the relation between $a,b,c$ which are in A.P is $b-a=c-b\Rightarrow 2b=a+c$. From this, we can conclude that $a+c$ is even as $2b$is always an even number. For the number $a+c$ to be even, we know that the condition is that the numbers $a,c$ are either even or odd. We can figure out that there are 250 odd numbers and 250 even numbers first 500 integers. The selection of $r$ objects from $n$ objects can be done in ${}^{n}{{C}_{r}}$ ways. So, we can get the required number of ways by calculating the sum of the number of ways of selecting 2 odd numbers out of 250 odd numbers and 2 even numbers from 250 even numbers.

Complete step by step answer:
We are given a set of 500 tickets which are numbered from 1 to 500. We are asked to find the number of ways of selecting three tickets so that the numbers on the tickets are in Arithmetic Progression. We know that the relation between $a,b,c$ which are in A.P is $\begin{align}
  & b-a=c-b \\
 & 2b=a+c \\
\end{align}$.
Let us consider the term $2b$. We can infer that it is always an even number for any value of b. So, we can conclude that the term $a+c$ should also be an even number.
We know that for the sum of two numbers to be even, both the numbers either should be even or should be odd.
So, we have the possible cases as both the number $a,c$ are either even or odd.
The odd numbers in 1 to 500 are $1,3,5,.....,499.$
It is an arithmetic progression with a common difference of 2 and first term 1.
${{n}^{th}}$term of an A.P with the first term as $a$ and a common difference $d$ is given by the relation ${{T}_{n}}=a+\left( n-1 \right)d$
We have the values as
$\begin{align}
  & {{T}_{n}}=499 \\
 & a=1 \\
 & d=2 \\
\end{align}$
We should find the value of n. Using the above relation, we get
$\begin{align}
  & 499=1+\left( n-1 \right)2 \\
 & 498=\left( n-1 \right)2 \\
 & n-1=\dfrac{498}{2}=249 \\
 & n=250 \\
\end{align}$
So, we have 250 odd numbers out of 500. The remaining are even numbers. SO, we have 250 odd and 250 even numbers in 1 to 500.
We know that the selection of $r$ objects from $n$ objects can be done in ${}^{n}{{C}_{r}}$ways.
Number of ways of selecting $a,c$ out of 250 odd numbers where $n=250,r=2$ is
Ways$={}^{250}{{C}_{2}}$
For this selection, we can infer that b is automatically selected with the selection of $a,c$.
Similarly
Number of ways of selecting $a,c$ out of 250 even numbers where $n=250,r=2$ is
Ways$={}^{250}{{C}_{2}}$
For this selection, we can infer that b is automatically selected with the selection of $a,c$.
We can infer that the total number of required ways is the sum of the two combinations done above.
Total ways
$\begin{align}
  & {}^{250}{{C}_{2}}+{}^{250}{{C}_{2}}=\dfrac{250!}{2!\left( 250-2 \right)!}+\dfrac{250!}{2!\left( 250-2 \right)!}=\dfrac{250!}{2!\left( 248 \right)!}+\dfrac{250!}{2!\left( 248 \right)!} \\
 & =\dfrac{250\times 249}{2}+\dfrac{250\times 249}{2}=250\times 249=62250 \\
\end{align}$
$\therefore $ The total number of required ways is 62250. The answer is option-C.

Note:
Students can confuse between the selection and permutation in this question. The progression is different based on the values of $a,c$ and students might use permutations instead of combinations. This confusion leads to an answer of ${}^{250}{{P}_{2}}+{}^{250}{{P}_{2}}=124500$. They might choose option-A. The trick is that even though the progression changes, the overall selection remains the same. So, we should understand and apply the correct formulae when it comes to permutations and combinations.