Question

If ${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}$ and ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}$ then $\dfrac{{{t}_{n}}}{{{S}_{n}}}$ is equal to
1. $\dfrac{n}{2}$
2. $\dfrac{n}{2}-1$
3. $n-1$
4. $\dfrac{2n-1}{2}$

Answer 1

To solve this question firstly you should have the idea of a combination. In the equation of ${{t}_{n}}$ add and subtract by $n$ to simplify the equation. After that apply the rule of combination and try to substitute the values from question to the simplified equation. When you apply all these steps you will get the final answer.

Complete step-by-step solution:
Firstly let us try to understand the concept of combination. It is a method of selecting items and the order of the items does not affect the answer that means order of items does not matter in combination. Whereas in the permutation is the method of arranging the items. And in permutation order of the items affect the overall answer.
The formula to solve the combination is
$^{n}{{C}_{r}}=\dfrac{n!}{(r)!(n-r)!}$
Where $C$ is the symbol used for combination.
From the given question we can say that,
${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}$
Now add and subtract by $n$in the numerator, by doing this there will be no change in the equation but it will be very helpful to further solve the equation. After applying this step we will get,
$\Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n-n+r}{{}^{n}{{C}_{r}}}}$
$\Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n-(n-r)}{{}^{n}{{C}_{r}}}}$
Simplifying the above expression, we will get
$\Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{{}^{n}{{C}_{r}}}-}\sum\limits_{r=0}^{n}{\dfrac{n-r}{{}^{n}{{C}_{r}}}}$ $........(1)$
By the concept of combination we know that $n-r=r$, let us prove this statement so that we can further use this for making the question simpler.
$^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$ and
$^{n}{{C}_{n-r}}=\dfrac{n!}{(n-n+r)!(n-r)!}$
$^{n}{{C}_{n-r}}=\dfrac{n!}{(r)!(n-r)!}$
From above we can say that both the equations will give the same answer. So we can say that $n-r=r$.
So substitute this value in equation $(1)$, we will get
$\Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{{}^{n}{{C}_{r}}}-}\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}$
In the first term of the equation $n$ is a constant value because summation function is applied on $r$so it will come out the summation function, we will get
$\Rightarrow {{t}_{n}}=n\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}-}\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}$ $........(2)$
And it is given in the question that
${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}$ , ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}$
So, substitute this value in equation $(2)$, we will get
$\Rightarrow {{t}_{n}}=n{{S}_{n}}-{{t}_{n}}$
$\Rightarrow 2{{t}_{n}}=n{{S}_{n}}$
$\Rightarrow \dfrac{{{t}_{n}}}{{{S}_{n}}}=\dfrac{n}{2}$
From the above calculation it proves that the ratio of ${{t}_{n}}$and ${{S}_{n}}$is $\dfrac{n}{2}$ .
Hence at last we can conclude that option $(1)$ is correct.

Note: Permutation and combination is a very useful concept in day to day life. To find the number of straight lines through $n$points we will simply apply the formula of combination i.e. $^{n}{{C}_{2}}$ and to find the number of triangles passing through $n$points we will apply the formula i.e. $^{n}{{C}_{3}}$.