Question

The value of $^{10}{{C}_{4}}{{+}^{10}}{{C}_{5}}$ is equal to
$\begin{aligned} & a)462 \\\ & b)466 \\\ & c)469 \\\ & d)465 \\\ \end{aligned}$

Answer 1

Now we know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ where n! = n × (n – 1) × (n – 2) × …. 3 × 2 × 1.
Hence using this we will find the value of $^{10}{{C}_{4}}$ and $^{10}{{C}_{5}}$ . Now we will add the following values. To find the value of $^{10}{{C}_{4}}{{+}^{10}}{{C}_{5}}$ .

Complete step by step answer:
Now first let us understand the meaning of the term $^{n}{{C}_{r}}$ . $^{n}{{C}_{r}}$ is formula which gives us number of ways to select r objects from n objects. For example we have 3 balls and we want to select 2 balls out of it. Then the number of ways in which we can select 2 balls is given by $^{3}{{C}_{2}}$ .
Now let us understand how to calculate the value of $^{n}{{C}_{r}}$ .
To do so we must first understand the term factorial.
Now n factorial is represented as n! and the value of n! is given by n × (n – 1) × (n – 2) × …. 3 × 2 × 1.
For example 4! = 4 × 3 × 2 × 1 = 24.
Now the value of $^{n}{{C}_{r}}$ is given by $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ .
Now let us consider $^{10}{{C}_{4}}$
$\begin{aligned} & ^{10}{{C}_{4}}=\dfrac{10!}{\left( 10-4 \right)!4!} \\\ & {{\Rightarrow }^{10}}{{C}_{4}}=\dfrac{10!}{6!4!} \\\ & {{\Rightarrow }^{10}}{{C}_{4}}=\dfrac{10\times 9\times 8\times 7\times 6!}{6!4!} \\\ & {{\Rightarrow }^{10}}{{C}_{4}}=\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2} \\\ \end{aligned}$
$\begin{aligned} & {{\Rightarrow }^{10}}{{C}_{4}}=10\times 3\times 7=210 \\\ & {{\therefore }^{10}}{{C}_{4}}=210............................\left( 1 \right) \\\ \end{aligned}$
Now consider the term $^{10}{{C}_{5}}$

& ^{10}{{C}_{5}}=\dfrac{10!}{\left( 10-5 \right)!5!} \\\ & {{\Rightarrow }^{10}}{{C}_{5}}=\dfrac{10\times 9\times 8\times 7\times 6\times 5!}{5!5!} \\\ & {{\Rightarrow }^{10}}{{C}_{5}}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1} \\\ & {{\Rightarrow }^{10}}{{C}_{5}}=2\times 3\times 7\times 6 \\\ & {{\therefore }^{10}}{{C}_{5}}=252......................\left( 2 \right) \\\ \end{aligned}$$ Now adding equation (1) and equation (2) we get, $\begin{aligned} & ^{10}{{C}_{4}}{{+}^{10}}{{C}_{5}}=252+210 \\\ & {{\therefore }^{10}}{{C}_{4}}{{+}^{10}}{{C}_{5}}=462 \\\ \end{aligned}$ Hence the value of $^{10}{{C}_{4}}{{+}^{10}}{{C}_{5}}$ is 462. **So, the correct answer is “Option a”.** **Note:** Now note that we have a property which says $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r+1}}{{=}^{n+1}}{{C}_{r+1}}$ hence using this we can say that $^{10}{{C}_{4}}{{+}^{10}}{{C}_{5}}{{=}^{11}}{{C}_{5}}$ and hence solve the equation easily.