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Question: If we have \[{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}\], then \(n\) is equal to A. 12 B. 26 C. 6 D....

If we have nC12=nC8{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}, then nn is equal to
A. 12
B. 26
C. 6
D. 20

Explanation

Solution

We first discuss the general form of combination and its general meaning with the help of variables. We express the mathematical notion with respect to the factorial form of nCr=n!r!×(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}. We try to find the value of nCn−r{}^{n}{{C}_{n-r}}. We complete the multiplication and find the relation.

Complete step-by-step solution:
We first try to find the general form of combination and its general meaning and then we put the values to find the solution.
The general form of combination is nCr{}^{n}{{C}_{r}}. It’s used to express the notion of choosing r objects out of n objects. The value of nCr{}^{n}{{C}_{r}} expresses the number of ways the combination of those objects can be done.
The simplified form of the mathematical expression nCr{}^{n}{{C}_{r}} is nCr=n!r!×(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}.
Here the term n!n! defines the notion of multiplication of first n natural numbers.
This means n!=1×2×3×....×nn!=1\times 2\times 3\times ....\times n.
The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation.
We now try to find the value of nCn−r{}^{n}{{C}_{n-r}} which is nCn−r=n!(n−r)!×[n−(n−r)]!=n!(n−r)!×r!{}^{n}{{C}_{n-r}}=\dfrac{n!}{\left( n-r \right)!\times \left[ n-\left( n-r \right) \right]!}=\dfrac{n!}{\left( n-r \right)!\times r!}.
Therefore, nCr=n!r!×(n−r)!=nCn−r{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}.
For nC12=nC8{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}, we can say n=12+8=20n=12+8=20. The correct option is D.

Note: There are some constraints in the form of nCr=n!r!×(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}. The general conditions are n≥r≥0;n≠0n\ge r\ge 0;n\ne 0. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing (n−r)\left( n-r \right) objects out of n objects. The mathematical expression is nCr=n!r!×(n−r)!=nCn−r{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}.