Question

Find the numerical value of

a.6! b.$\left(\begin{array}{c}10\\ 9\end{array}\right)$c. $\left(\begin{array}{c}10\\ 1\end{array}\right)$d.$\left(\begin{array}{c}6\\ 3\end{array}\right)$e.0!

Short answer

The results for the given numerical values are as follows:

1. 720
2. 10
3. 10
4. 20
5. 1

Step-by-step solution

Important formula

The formula for combination is ${}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}$.

Finding the value of 6!

a.

$\begin{array}{l}6!=6\times 5\times 4\times 3\times 2\times 1\\ 6!=720\end{array}$.

Therefore, the value of 6! is 720.

Step 3:Finding the value of (109)

b.

$$\begin{gathered} {}^{10}C_{19}=\frac{10!}{9!\left(10-9\right)!} \\ =\frac{10!}{9!\left(1\right)!} \\ =10 \end{gathered}$$

Accordingly, the value is 10.

Evaluate the value of (101)

c.

$$\begin{gathered} {}^{10}C_1=\frac{10!}{1!\left(10-1\right)!} \\ =\frac{10!}{9!\left(1\right)!} \\ =10 \end{gathered}$$

So, the required value is 10.

Finding the value of (63)

d.

$$\begin{gathered} {}^{6}C_3=\frac{6!}{3!\left(6-3\right)!} \\ =\frac{6!}{3!\left(3\right)!} \\ =20 \end{gathered}$$

Hence, the value is 20.

Finding the value of 0!

The value of 0! is 1.