Question

If n objects are arranged in a row, then the number of ways of selecting three out of these objects so that no two of them are next to each other is:
(A). $\dfrac{\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)}{6}$
(B). ${}^{n-2}{{C}_{3}}$
(C). ${}^{n-3}{{C}_{3}}+{}^{n-3}{{C}_{2}}$
(D). None of these

Answer 1

Hint: Find randomly selects any 3 objects. Now keep the 4 sets of numbers of objects as 4 variables. Now we know that the sum of all these 4 variables is total -3. -3 is written because we already selected 3 objects. Now we want a selection such that number 2 is together. So, we get the condition that two variables are $\ge 1$ and the remaining two are $\ge 0$. Now write the $\ge 1$ variables as variable -1 which will also be $\ge 0$. Now we have a sum of 4 terms that is equal to a value, where all terms are $\ge 0$. Now by general knowledge of combinatory, we know number of solutions of $a+b=c+d=n$ is ${}^{n+r-1}{{C}_{n}}$ . where ‘r’ is the number of variables.

Complete step-by-step solution -
Combinations: - It is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in the combination you can select items in any order. Formula is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Given condition in the question can be written as:
 Select 3 objects from n objects in a row. Such that no two are together.
 Now, let the n objects in a row are represented as –
$-\square -\bigcirc -\Delta -$
The objects square, circle, triangle represent the objects that are selected.
Now the condition says that the objects between (square, circle); (circle, triangle) must be greater than 0.
Now, let us assume the number of objects before the square is “a”.
Next, let us assume the number of objects between the square, the circle is b.
Then let us assume the number of objects between the circle, the triangle is “c”.
Next, let us assume the number of objects after the triangle as d.
As we already removed 3 objects out of n, we get sum of these as, $a+b+c+d=n-3$ .
As per our condition, we may take zero objects at the start, end but not in between. So, we say: $a,d\ge 0$ $b,c\ge 1$ .
If we subtract 1 from b, c. We can get inequality as $b-1,c-1\ge 0$ .
So, if we subtract 2 from both sides of the equation, we get $a+b+c+d-2=n-5$ .
Here, $a,b-1,c-1,d\ge 0$ .
By combination $x+y+z=r$ number of ways ${}^{r+3-1}{{C}_{r}}$.
So, here by substituting these values, we get ${}^{\left( n-5 \right)+4-1}{{C}_{n-5}}$.
By simplifying, we get the value as: ${}^{n-2}{{C}_{n-5}}$.
We know that ${}^{n}{{c}_{r}}={}^{n}{{c}_{n-r}}$. Using this we get ${}^{n-2}{{C}_{n-2-\left( n-5 \right)}}={}^{n-2}{{C}_{3}}$.
So, the number of ways for a given condition is ${}^{n-2}{{C}_{3}}$ -(1).
We know that ${}^{n+1}{{C}_{r}}={}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}$. Using this we get ${}^{n-2}{{C}_{3}}={}^{n-3}{{C}_{3}}+{}^{n-3}{{C}_{2}}$ -(2).
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$. We use these results for the value we obtained in equation (1).
So, we get ${}^{n-2}{{C}_{3}}=\dfrac{\left( n-2 \right)!}{3!\left( n-2-3 \right)!}$.
$\Rightarrow {}^{n-2}{{C}_{3}}=\dfrac{\left( n-2 \right)!}{3!\left( n-5 \right)!}$.
$\Rightarrow {}^{n-2}{{C}_{3}}=\dfrac{\left( n-2 \right)\times \left( n-3 \right)\times \left( n-4 \right)\times \left( n-5 \right)\times \left( n-6 \right)\times ......\times 2\times 1}{\left( 3\times 2\times 1 \right)\times \left( \left( n-5 \right)\times \left( n-6 \right)\times ......\times 2\times 1 \right)}$.
$\Rightarrow {}^{n-2}{{C}_{3}}=\dfrac{\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)}{6}$ -(3).
From equations (1), (2) and (3), we have found the correct options as (a), (b) and (c).

Note: The important idea here is to convert the problem into the mathematical sum, whenever you see them like no 2 terms together you can use this idea. Remember all the variables must be $\ge 0$ if not convert the variable into that form to apply the formula. While changing the variable be -1 don’t forget to apply on R.H.S. We should not stop solving the problem once we find the result as ${}^{n-2}{{C}_{3}}$. Because we can expect multiple correct options for the problems given sometimes.