Question

What is the formula of $${}^n{C}_n$$?

Answer 1

Hint: In permutation, selection is made but beyond selection, order or arrangement is important. In combination is a way of selecting items from a collection where the order of selection or arrangement does not matter. We will use the combination formula and then substitute in place of r. Also we know that $$0!=1$$. Hence, using all these we will get the final formula of $${}^n{C}_n$$ .

Complete step-by-step answer:
We have to find the formula of $${}^n{C}_n$$.
We know the formula of combination as:
$${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$
Here, n represents the number of items, and r represents the number of items being chosen at a time.
Thus, for finding the given value $${}^n{C}_n$$ :
$${}^n{C}_n={\displaystyle \frac{n!}{n!(n-n)!}}$$
On evaluating this above equation, we will get,
$$\Rightarrow {}^n{C}_n={\displaystyle \frac{n!}{n!(0)!}}$$
$$\Rightarrow {}^n{C}_n={\displaystyle \frac{n!}{n!(1)}}$$
As we know that, $$0!=1$$ and so applying this we will get,
$$\Rightarrow {}^n{C}_n={\displaystyle \frac{n!}{n!}}$$
$$\Rightarrow {}^n{C}_n=1$$
Hence, the formula of $${}^n{C}_n=1$$.
In short,
$$\therefore {}^n{C}_n$$
$$={}^n{C}_{n-n}$$
$$={}^n{C}_0$$
$$=1$$
So, the correct answer is “1”.

Note: The ways of arranging or selecting a smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection are called permutations. Permutation often occurs when different orderings on certain finite sets are considered. The number of ways in which a smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important are called combinations. In smaller cases, it is possible to count the number of combinations. A permutation is used for the list of data where the order of the data matters and the combination is used for a group of data where the order of data doesn’t matter. The formulas are both are:
1) Permutation: $${}^n{P}_r={\displaystyle \frac{n!}{(n-r)!}}$$
2) Combination: $${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$