Question

Evaluate ${}_{5}{{C}_{3}}$.

Answer 1

Hint: The combination of n things taken r at a time is given by ${}_{n}{{C}_{r}}$. The given question asks us to find out the combinations of 5 things taken 3 at a time. We use the formula of ${}_{n}{{C}_{r}}$ that is $\dfrac{n!}{\left( n-r \right)!\text{ }r!}$ where n =5 and r=3 as per the question to get the required result.

Complete step by step answer:
Combinations usually are the number of ways the objects can be selected without replacement.
Order is not usually a constraint in combinations, unlike permutations.
The Combination of n things or items taken r at a time is denoted by ${}_{n}{{C}_{r}}$.
n = no of things
r = no of things taken at the time.
The combination of n things taken r at a time or ${}_{n}{{C}_{r}}$ is given by the formula
 $\Rightarrow {}_{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\text{ }r!}$
The exclamation mark in the above formula denotes factorial.
Factorial of a number in mathematics is the product of all the positive numbers less than or equal to a number.
The multiplication happens to a given number down to the number one or till the number one is reached.
Example: Factorial of n is n! and the value of n! is $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\ldots \ldots \ldots 1$
In the given question,
We need to evaluate ${}_{5}{{C}_{3}}$
We are supposed to find the combination of 6 things taken 3 at a time.
 $\Rightarrow {}_{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\text{ }r!}$
Here,
n = 5
r = 3
Substituting the values,
$\Rightarrow {}_{5}{{C}_{3}}=\dfrac{5!}{\left( 5-3 \right)!\text{ 3}!}$
$\Rightarrow {}_{5}{{C}_{3}}=\dfrac{5!}{2!\text{ 3}!}$
Writing the numerator as the product of prime factors,
$\Rightarrow {}_{5}{{C}_{3}}=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}$
$\Rightarrow {}_{5}{{C}_{3}}=\dfrac{5\times 2}{1}$
$\Rightarrow {}_{5}{{C}_{3}}=10$

The combination of 5 things taken 3 at a time or the value of ${}_{5}{{C}_{3}}$ is 10

Note: We need to know the basic concepts and definitions of combinations to solve this problem. Combinations usually are the number of ways the objects can be selected without replacement. Common mistakes like double counting, confusion of values should be avoided to get precise results.