Question

Find the value of \[{}^{10}{{C}_{5}}+2\cdot \left( {}^{10}{{C}_{4}} \right)+{}^{10}{{C}_{3}}\].

Answer 1

Hint: In this question, by using the laws of binomial coefficients we can rewrite the given equation and simplify each term accordingly to get the solution.
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\]

Complete step-by-step answer:
Binomial Theorem for Positive Integer:
If n is any positive integer, then
\[{{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+......{}^{n}{{C}_{n}}{{a}^{n}}\]
\[{{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{a}^{r}}\]
This is called binomial theorem.
Here,\[{}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}}etc\]are called binomial coefficients and
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\] for \[0\le r\le n\]
Important result on Binomial Coefficients:
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\]
From the given equation in the question we get,
\[\Rightarrow {}^{10}{{C}_{5}}+2\cdot \left( {}^{10}{{C}_{4}} \right)+{}^{10}{{C}_{3}}\]
Now, this equation can also be written as:
\[\Rightarrow {}^{10}{{C}_{5}}+{}^{10}{{C}_{4}}+{}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}\]
As we already know that.
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\]
Now, by using the result on the binomial coefficients we can write it as.
\[\Rightarrow {}^{10+1}{{C}_{5}}+{}^{10+1}{{C}_{4}}\]
Which can be further written as.
\[\Rightarrow {}^{11}{{C}_{5}}+{}^{11}{{C}_{4}}\]
Now, by again applying the result of the binomial coefficients we get,
\[\Rightarrow {}^{11+1}{{C}_{5}}\]
This on further simplification can be written as.
\[\Rightarrow {}^{12}{{C}_{5}}\]
Now, let us find this value by using the formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
Let us now substitute the corresponding values in the above formula.
\[\Rightarrow {}^{12}{{C}_{5}}=\dfrac{12!}{\left( 12-5 \right)!5!}\]
This on further simplification we get,
\[\Rightarrow {}^{12}{{C}_{5}}=\dfrac{12!}{7!5!}\]
Now on expanding the factorial terms in the numerator and denominator we get,
\[\Rightarrow {}^{12}{{C}_{5}}=\dfrac{12\times 11\times 10\times 9\times 8\times 7!}{7!5!}\]
Now, on cancelling the common terms in the numerator and the denominator we get,
\[\Rightarrow {}^{12}{{C}_{5}}=\dfrac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow {}^{12}{{C}_{5}}=12\times 11\times 3\times 2 \\
 & \therefore {}^{12}{{C}_{5}}=792 \\
\end{align}\]
Hence, the value of the given function \[{}^{10}{{C}_{5}}+2\cdot \left( {}^{10}{{C}_{4}} \right)+{}^{10}{{C}_{3}}\] is 792.


Note: Instead of using the result of the binomial coefficients we can also directly expand it by using the coefficient formula and then add them all together but it will be a lengthy process. Both the processes give the same result.
While calculating we need to be careful and should not miss any of the terms because neglecting one of the terms changes the result completely. When expanding the factorial terms we need to simplify them by cancelling the common terms and should not miss any terms while expanding.