Question

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?

Answer 1

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’.

Complete step by step answer:

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.