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# If one-quarter of all three-element subsets of the set ${\text{A = }}{{\text{a}}_{\text{1}}}{\text{,}}{{\text{a}}_{\text{2}}}{\text{,}}{{\text{a}}_{\text{3}}}....{{\text{a}}_{\text{n}}}$ contains the element ${{\text{a}}_{\text{3}}}$, then n is equal to?

Hint: First we’ll find the number of subsets of A containing three elements and the number of subsets of A containing three elements in which one of the elements is sure to be ${{\text{a}}_{\text{3}}}$. Then will form an equation according to the data given in the question, we’ll obtain an equation in ‘n’ which will get us the value of ‘n’.

Given data: one-quarter of all three-element subsets of the set ${\text{A = }}{{\text{a}}_{\text{1}}}{\text{,}}{{\text{a}}_{\text{2}}}{\text{,}}{{\text{a}}_{\text{3}}}....{{\text{a}}_{\text{n}}}$ contains the element ${{\text{a}}_{\text{3}}}$
Number of subsets of A containing three elements is ${}^{\text{n}}{{\text{C}}_{\text{3}}}$
The number of subsets of A containing three elements in which one of the elements is sure to be ${{\text{a}}_{\text{3}}}$is${}^{{\text{n - 1}}}{{\text{C}}_{\text{2}}}$
Now, according to the given statement, we can say that
$\dfrac{{\text{1}}}{{\text{4}}}{}^{\text{n}}{{\text{C}}_{\text{3}}}{\text{ = }}{}^{{\text{n - 1}}}{{\text{C}}_{\text{2}}}$
Using ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n - r}}} \right){\text{!}}}}$ , we’ll get
$\dfrac{{\text{1}}}{{\text{4}}}\dfrac{{{\text{n!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{2!}}\left( {{\text{n - 1 - 2}}} \right){\text{!}}}}$
Now, using ${\text{n! = n}}\left( {{\text{n - 1}}} \right){\text{!}}$
$\dfrac{{\text{1}}}{{\text{4}}}\dfrac{{{\text{n(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{2!}}\left( {{\text{n - 3}}} \right){\text{!}}}} \\ \Rightarrow \dfrac{{\text{1}}}{{\text{4}}}\dfrac{{\text{n}}}{{{\text{3!}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{2!}}}} \\$
Now, solving for n, we’ll obtain
${\text{n = }}\dfrac{{\text{4}}}{{{\text{2!}}}}{\text{3!}} \\ \Rightarrow {\text{n = }}\dfrac{{\text{4}}}{{{\text{2!}}}}{\text{3(2!)}} \\ \Rightarrow {\text{n = 4(3)}} \\ \Rightarrow {\text{n = 12}} \\$
Therefore, the value of n is 12

Note: We can also find the number of subsets of A containing three elements in which one of the elements is sure to be ${{\text{a}}_{\text{3}}}$that will be equal to the difference between the number of subsets of A containing three elements and the number of three-element subsets of A do not contain ${{\text{a}}_{\text{3}}}$ i.e.
${\text{ = }}{}^{\text{n}}{{\text{C}}_{\text{3}}}{\text{ - }}{}^{{\text{n - 1}}}{{\text{C}}_{\text{3}}}$
Using ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n - r}}} \right){\text{!}}}}$ ,
${\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ - }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 4}}} \right){\text{!}}}}$
Using ${\text{n! = n}}\left( {{\text{n - 1}}} \right){\text{!}}$
${\text{ = }}\dfrac{{{\text{n(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ - }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 4}}} \right){\text{!}}}}$
Dividing and multiplying the second term with (n-3)
${\text{ = }}\dfrac{{{\text{n(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ - }}\left[ {\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 4}}} \right){\text{!}}}}\left( {\dfrac{{{\text{n - 3}}}}{{{\text{n - 3}}}}} \right)} \right]$
Using ${\text{n}}\left( {{\text{n - 1}}} \right){\text{! = n!}}$
${\text{ = }}\dfrac{{{\text{n(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{ - }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}\left( {{\text{n - 3}}} \right)$
Now, taking common $\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}$from both terms
${\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}\left[ {{\text{n - }}\left( {{\text{n - 3}}} \right)} \right]$
On further simplification we get,
${\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{3!}}\left( {{\text{n - 3}}} \right){\text{!}}}}{\text{3}}$
Using ${\text{n}}\left( {{\text{n - 1}}} \right){\text{! = n!}}$, and simplifying we get,
${\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{2!}}\left( {{\text{n - 3}}} \right){\text{!}}}}$
${\text{ = }}\dfrac{{{\text{(n - 1)!}}}}{{{\text{2!}}\left( {{\text{n - 1 - 2}}} \right){\text{!}}}}$
We can write it as ${}^{{\text{n - 1}}}{{\text{C}}_{\text{2}}}$and is having the same value as in our above solution.