Question

Verify that ${}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}}$ .

Answer 1

Hint: We first need to write down the formula for ${}^{n}{{C}_{r}}$ which is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Then, we need to put values $n=10,10,11$ and $r=4,3,4$ respectively to find the values of ${}^{10}{{C}_{4}},{}^{10}{{C}_{3}},{}^{11}{{C}_{4}}$ . Having done so, we then need to check the value of the LHS and see if it equals the RHS. If yes, then our expression is verified.

Complete step-by-step solution:
The expression that we are given in this problem is,
${}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}}$
We can prove the expression by equating the LHS and the RHS of the equation. But, before that, we need to simplify each of the sides to a natural number each and then show that the two natural numbers are one and the same.
We know that the formula for ${}^{n}{{C}_{r}}$ is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Now, we need to put some values in this formula one by one to find the values of ${}^{10}{{C}_{4}},{}^{10}{{C}_{3}},{}^{11}{{C}_{4}}$ . At first, putting $n=10,r=3$ , we get,
$\begin{align}
  & \Rightarrow {}^{10}{{C}_{3}}=\dfrac{10!}{3!\left( 10-3 \right)!} \\
 & \Rightarrow {}^{10}{{C}_{3}}=\dfrac{10!}{3!\times 7!}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120 \\
\end{align}$
Then, putting $n=10,r=4$ , we get,
$\begin{align}
  & \Rightarrow {}^{10}{{C}_{4}}=\dfrac{10!}{4!\left( 10-4 \right)!} \\
 & \Rightarrow {}^{10}{{C}_{4}}=\dfrac{10!}{4!\times 6!}=\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=210 \\
\end{align}$
The, putting $n=11,r=4$ , we get,
$\begin{align}
  & \Rightarrow {}^{11}{{C}_{4}}=\dfrac{11!}{4!\left( 11-4 \right)!} \\
 & \Rightarrow {}^{11}{{C}_{4}}=\dfrac{11!}{4!\times 7!}=\dfrac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}=330 \\
\end{align}$
The LHS becomes,
$\Rightarrow {}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}=120+210=330$
And, the RHS becomes,
$\Rightarrow {}^{11}{{C}_{4}}=330$
This means that the left hand side has become equal to the right hand side. This means that the expression is verified.
Thus, we can conclude that ${}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}}$ .

Note: We can also solve the problem in another way. There is a property of combinations which goes as ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ . We just simply need to check whether this relation is valid for this problem or not. By simple comparison, we can see that in our problem, $n=10,r=4$ . So, the property is applicable for our problem and thus the expression is verified.