
Evaluate the expression ${}^{2}{{C}_{2}}$
Answer
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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 2 things taken 2 at a time. We use the formula of ${}^{n}{{C}_{r}}$ that is $\dfrac{n!}{\left( n-r \right)!\text{ }r!}$ where n =2 and r=2 as per the question to get the required result.
Complete step-by-step solution:
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
$ {}^{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 ${}^{2}{{C}_{2}}$
We are supposed to find the combination of 2 things taken 2 at a time.
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\text{ }r!}$
Here,
n = 2
r = 2
Substituting the values,
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2!}{\left( 2-2 \right)!\text{ 2}!}$
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2!}{0!\text{ 2}!}$
Writing the numerator as the product of prime factors,
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2\times 1}{2\times 1}$
$\Rightarrow {}^{2}{{C}_{2}}=1$
There is only one way to find the combination of two things taken two at a time.
The combination of 2 things taken 2 at a time or the value of ${}^{2}{{C}_{2}}$ is 1
Note: Combinations determine all the possible selections in a set of items or things. The factorial of a number n should only go down to 1 and not zero. The order of selection or sequence does not matter in the case of combinations, unlike permutations.
Complete step-by-step solution:
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
$ {}^{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 ${}^{2}{{C}_{2}}$
We are supposed to find the combination of 2 things taken 2 at a time.
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\text{ }r!}$
Here,
n = 2
r = 2
Substituting the values,
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2!}{\left( 2-2 \right)!\text{ 2}!}$
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2!}{0!\text{ 2}!}$
Writing the numerator as the product of prime factors,
$\Rightarrow {}^{2}{{C}_{2}}=\dfrac{2\times 1}{2\times 1}$
$\Rightarrow {}^{2}{{C}_{2}}=1$
There is only one way to find the combination of two things taken two at a time.
The combination of 2 things taken 2 at a time or the value of ${}^{2}{{C}_{2}}$ is 1
Note: Combinations determine all the possible selections in a set of items or things. The factorial of a number n should only go down to 1 and not zero. The order of selection or sequence does not matter in the case of combinations, unlike permutations.
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