Question

How do you evaluate ${}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}$ ?

Answer 1

Hint: A permutation or combination is a set of ordered things. From the symbol ‘C’ we get to know that it is a combination. If you don’t care about the order of the things then it’s a combination means combination is the selection of items where order doesn’t matter and it is used for selection of items. A combination of $n$ distinct objects taken $r$ at a time is a selection of$r$ objects out of these n objects$\left( 0\le r\le n \right)$. Then, the total number of different combinations of $n$ distinct objects taken $r$ at a time is denoted by ${}^{n}{{C}_{r}}$ . It has other symbols also as $C_{r}^{n}$or${}^{n}{{C}_{r}}$or$\left( \underset{r}{\overset{n}{\mathop {}}}\, \right)$ which is binomial coefficient or $C\left( n,r \right)$ . It’s general formula is
If repetition is allowed $\dfrac{n!}{r!\left( n-r \right)!}$
If repetition is not allowed $\dfrac{\left( n+r-1 \right)!}{r!\left( n-1 \right)!}$

Complete step by step solution:
To calculate a combination, you will need to calculate a factorial.
$\Rightarrow n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)................3\cdot 2\cdot 1$
The combination general formula is,
${{\Rightarrow }^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Where,
$n=$ population/ total objects in the set
$r=$ picks/selected objects from the set
Given combination are ${}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}$
We calculate it partially,
First,${}^{8}{{C}_{5}}$ here $n=8$ and $r=5$
\[\begin{align}
  & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{8!}{5!\left( 8-5 \right)!} \\
 & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{8!}{5!\cdot 3!} \\
 & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{8\times 7\times 6\times 5!}{5!\times 3\times 2\times 1} \\
\end{align}\]
$5!$ is cancelled out then we have,
$\begin{align}
  & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{8\times 7\times 6}{6} \\
 & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{336}{6}=56 \\
\end{align}$
After dividing it we have$^{8}{{C}_{5}}=56$
Similarly we calculate ${}^{7}{{C}_{3}}$ here $n=7$ and $r=5$
${{\Rightarrow }^{7}}{{C}_{3}}=\dfrac{7!}{3!\left( 7-3 \right)!}=\dfrac{7!}{3!\centerdot 4!}=\dfrac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}$
Now 4! is cancelled out then we have,
${{\Rightarrow }^{7}}{{C}_{3}}=\dfrac{7\times 6\times 5}{6}$
Now 6 is cancelled out then we have,
$\Rightarrow $ $7\times 5=35$
So, we have ${}^{7}{{C}_{3}}$=35
According to the question ${}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}$ we have to multiply it then we have,
$\Rightarrow {}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}=56\times 35=1960$

$1960$ is our answer after solving this combination.

Note: Firstly, we have to write the formula correctly which is $\dfrac{n!}{r!\left( n-r \right)!}$. Do not confuse taking $n$ and $r$,$c\left( n,r \right)$ here $n$ is the number of items in the set and $r$ is the number of items selected from the set. Avoid multiplying like ${}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}={}^{56}{{C}_{15}}$. Do not overlap and miss any number during multiplication.