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Question: How do you find the binomial coefficient of \[^{12}{C_0}\] ....

How do you find the binomial coefficient of 12C0^{12}{C_0} .

Explanation

Solution

On order to solve this question we need to apply the binomial expansion then there is the formula according to which we will find the value of 12C0^{12}{C_0} and the formula of the expansion is nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} .
According to this formula we will up to the final answer.

Complete step by step solution:
For solving this question we will have to know what is the binomial expansion and what are the formula to expand binomial expansion so we will learn it first;

Binomial Theorem – As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial Theorem.

The Binomial Theorem is the method of expanding an expression which has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc.

Now we will know what is a binomial expression;

Binomial expression: A binomial expression is an algebraic expression which contains two dissimilar terms.

Formula of binomial expression:
nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}

So by applying the formula for the given question;
12C0=12!0!(12−0)!^{12}{C_0} = \dfrac{{12!}}{{0!(12 - 0)!}}

An important thing should be noted, the value of 0!=10! = 1 ;

So putting this value in equation and on further solving it we will get;
12C0=12!12!^{12}{C_0} = \dfrac{{12!}}{{12!}}

Now finally we will get;
12C0=1^{12}{C_0} = 1

And hence the final answer is 1.

Note: Now we will show 0!=10! = 1 . Let n be a whole number, where n!n! is defined as the product of factors including n itself and everything below it. What it means is that you first start writing the whole number n then count down until you reach the whole number 1.

The general formula of factorial can be written in fully expanded form as
n!=n×(n−1)×(n−2)×(n−3)×.............×3×2×1n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ............. \times 3 \times 2 \times 1

or in partially expanded form as
n!=n(n−1)!n! = n(n - 1)!

We know with absolute certainty that 1!=11! = 1, where n=1. If we substitute that value of n into the second formula which is the partially expanded form of n!n! . we obtain the followingn!=n(n−1)!n! = n(n - 1)!

On putting the value n=1
1!=1(1−1)!1! = 1(1 - 1)!
on further solving
1=1(0)!1 = 1(0)!

Finally we get
0!=10! = 1