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Question: For n being a natural number, if \({}^{2n}{C_r} = {}^{2n}{C_{r + 2}}\), find \(r\). A. \(n\) B. ...

For n being a natural number, if 2nCr=2nCr+2{}^{2n}{C_r} = {}^{2n}{C_{r + 2}}, find rr.
A. nn
B. n−1n - 1
C. n−2n - 2
D. n−3n - 3

Explanation

Solution

Hint – We will start solving this question by noting down the given equation. We will be solving this question by using the formula of combination, i.e., nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}, then we will get the required value of rr.

Complete step-by-step answer:
We know that,
A combination is a selection of some or all of a number of different objects. When we select rr objects out of nn objects (r⩽n)\left( {r \leqslant n} \right) each solution is called a combination.
The combination of nn objects taken rr at a time is written as C(n,r)C\left( {n,r} \right) or nCr{}^n{C_r}.
The given equation is,
2nCr=2nCr+2{}^{2n}{C_r} = {}^{2n}{C_{r + 2}}
Now, we know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}, therefore the LHS of the above equation becomes,
2nC2n−r=2nCr+2{}^{2n}{C_{2n - r}} = {}^{2n}{C_{r + 2}}
We know that, as
2nC2n−r=2nCr+2{}^{2n}{C_{2n - r}} = {}^{2n}{C_{r + 2}}
Therefore,
2n−r=r+2 ⇒r+r=2n−2 ⇒2r=2n−2 ⇒2r=2(n−1) ⇒r=n−1  2n - r = r + 2 \\\ \Rightarrow r + r = 2n - 2 \\\ \Rightarrow 2r = 2n - 2 \\\ \Rightarrow 2r = 2\left( {n - 1} \right) \\\ \Rightarrow r = n - 1 \\\
Therefore, rr is n−1n - 1.
Hence, option B is the correct answer.

Note – A combination is that which determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, we can select the items in any order. These questions must be solved carefully and all the basic formulas must be remembered.