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Question: How do you evaluate \({}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}\) ?...

How do you evaluate 8C5⋅7C3{}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}} ?

Explanation

Solution

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 nn distinct objects taken rr at a time is a selection ofrr objects out of these n objects(0≤r≤n)\left( 0\le r\le n \right). Then, the total number of different combinations of nn distinct objects taken rr at a time is denoted by nCr{}^{n}{{C}_{r}} . It has other symbols also as CrnC_{r}^{n}ornCr{}^{n}{{C}_{r}}or(nr )\left( \underset{r}{\overset{n}{\mathop {}}}\, \right) which is binomial coefficient or C(n,r)C\left( n,r \right) . It’s general formula is
If repetition is allowed n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!}
If repetition is not allowed (n+r−1)!r!(n−1)!\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.
⇒n!=n(n−1)(n−2)(n−3)................3⋅2⋅1\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,
⇒nCr=n!r!(n−r)!{{\Rightarrow }^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Where,
n=n= population/ total objects in the set
r=r= picks/selected objects from the set
Given combination are 8C5⋅7C3{}^{8}{{C}_{5}}\centerdot {}^{7}{{C}_{3}}
We calculate it partially,
First,8C5{}^{8}{{C}_{5}} here n=8n=8 and r=5r=5

& {{\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{aligned}$$ $5!$ is cancelled out then we have, $\begin{aligned} & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{8\times 7\times 6}{6} \\\ & {{\Rightarrow }^{8}}{{C}_{5}}=\dfrac{336}{6}=56 \\\ \end{aligned}$ 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.