Find the value of ${}^{6}{{C}_{0}}{{.}^{12}}{{C}_{6}}{{-}^{6}}{{C}_{1}}{{.}^{11}}{{C}_{6}}{{+}^{6}}{{C}_{2}}{{.}^{10}}{{C}_{6}}{{-}^{6}}{{C}_{3}}{{.}^{9}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{8}}{{C}_{6}}{{-}^{6}}{{C}_{5}}{{.}^{7}}{{C}_{6}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{6}}$

Question

Find the value of ${}^{6}{{C}_{0}}{{.}^{12}}{{C}_{6}}{{-}^{6}}{{C}_{1}}{{.}^{11}}{{C}_{6}}{{+}^{6}}{{C}_{2}}{{.}^{10}}{{C}_{6}}{{-}^{6}}{{C}_{3}}{{.}^{9}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{8}}{{C}_{6}}{{-}^{6}}{{C}_{5}}{{.}^{7}}{{C}_{6}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{6}}$

Vedantu Content Team · Accepted Answer

Hint: Now to solve the expression we will first expand the whole equation by using the formula $\dfrac{n!}{r!\left( n-r \right)!}$ . Now once we have expanded we will simplify each term by cancelling denominator and numerator. Hence we will simplify the terms and find the value of the given expression.  Complete step by step solution:Now first let us understand the concept of the number $^{n}{{C}_{r}}$. To understand it we must first understand the factorial of a number. Now $n!$ is defined as $n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times ...\times 3\times 2\times 1$.Let us see an example. Consider we want to find the value of 5!. Hence we write it as 5! = 5 × 4 × 3 × 2 × 1 = 120. Hence the value of 5! = 120. The number n! gives the number of arrangements of n objects into n places.Now suppose we have n objects and we want to select r objects among them. Then the number of possible ways to do so is given by the number $^{n}{{C}_{r}}$ . Now the number $^{n}{{C}_{r}}$ is defined as $\dfrac{n!}{r!\left( n-r \right)!}$. Let us take an example to understand this.Suppose we have 5 different balls and we want to select 3 balls out of those then the number of ways to do so is $\dfrac{5!}{\left( 5-3 \right)!3!}=\dfrac{5!}{2!3!}=\dfrac{5\times 4\times 3!}{3!\times \left( 2 \right)}=10$ . Now consider the given series ${}^{6}{{C}_{0}}{{.}^{12}}{{C}_{6}}{{-}^{6}}{{C}_{1}}{{.}^{11}}{{C}_{6}}{{+}^{6}}{{C}_{2}}{{.}^{10}}{{C}_{6}}{{-}^{6}}{{C}_{3}}{{.}^{9}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{8}}{{C}_{6}}{{-}^{6}}{{C}_{5}}{{.}^{7}}{{C}_{6}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{6}}$Hence now let us expand the series by using the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . $$\dfrac{6!}{6!0!}\times \dfrac{12!}{6!6!}-\dfrac{6!}{1!5!}\times \dfrac{11!}{6!5!}+\dfrac{6!}{2!4!}\times \dfrac{10!}{6!4!}-\dfrac{6!}{3!3!}\times \dfrac{9!}{6!3!}+\dfrac{6!}{4!2!}\times \dfrac{8!}{6!2!}-\dfrac{6!}{5!1!}\dfrac{7!}{6!1!}+\dfrac{6!}{6!0!}\times \dfrac{6!}{6!} $$$$\Rightarrow \dfrac{1}{6!}\times \dfrac{12!}{6!}-\dfrac{1}{5!}\times \dfrac{11!}{5!}+\dfrac{1}{2!4!}\times \dfrac{10!}{4!}-\dfrac{1}{3!3!}\times \dfrac{9!}{3!}+\dfrac{1}{4!2!}\times \dfrac{8!}{2!}-\dfrac{1}{5!1!}\dfrac{7!}{1!}+1 $$$$\Rightarrow \dfrac{12\times 11...\times 8\times 7\times 6!}{6!\times 6!}-\dfrac{11\times 10\times ...\times 6\times 5!}{5!\times 5!}+\dfrac{10\times 9\times ...\times 6\times 5\times 4!}{2!\times 4!\times 4!}-\dfrac{9\times 8\times ...\times 4\times 3!}{3!3!3!}+\dfrac{8\times 7\times ...\times 4!}{2!2!4!}-\dfrac{7\times 6\times 5!}{5!} $$$$\Rightarrow \dfrac{12\times 11\times ...\times 8\times 7}{6!}-\dfrac{11\times 10\times 9\times 8\times 7\times 6}{5!}+\dfrac{10\times 9\times 8\times 7\times 6\times 5}{2!\times 4!}-\dfrac{9\times 8\times 7\times 6\times 5\times 4}{3!3!}+\dfrac{8\times 7\times 6\times 5}{2!2!}-\dfrac{7\times 6}{1} $$$$\Rightarrow \dfrac{12\times 11\times ...\times 8\times 7}{6\times 5\times 4\times 3\times 2}-\dfrac{11\times 10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}+\dfrac{10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 2}-\dfrac{9\times 8\times 7\times 6\times 5\times 4}{6\times 6}+\dfrac{8\times 7\times 6\times 5}{4}-\dfrac{42}{1} $$$$\Rightarrow \dfrac{2\times 11\times 3\times 2\times 7}{1}-\dfrac{11\times 2\times 9\times 2\times 7}{1}+\dfrac{10\times 9\times 7\times 5}{1}-\dfrac{3\times 4\times 7\times 5\times 4}{1}+\dfrac{2\times 7\times 6\times 5}{1}-\dfrac{42}{1}  $$$$\Rightarrow 924-2772+3150-1680+420-42 $$$$\Rightarrow 0 $$Hence the answer of the given equation is zero.Note: Note that here we have used 0! = 1. This is the standard value of 0!. Now to obtain this value we can consider that $n!=n\times \left( n-1 \right)!$ . Substituting n = 1 in the equation we get, 0! = 1. Hence the value of 0! is the same as 1! = 1.