Question

Evaluate \[^3{C_2}\]

Answer 1

Hint: A combination or permutation is a set of ordered things. Where \['C'\] represents combination and \['P'\] represents permutation. Combination is basically the number of different ways in which objects can be arranged without considering any order. It is represented as \[^n{C_r}\] .
Where, \[n = \] total number of objects, \[r = \] number of chosen objects, \[^n{C_r} = \] number of combinations.
Also the factorial in mathematics is the product of all positive integers less than or equal to a given positive integer, and is represented by the integer and the exclamation mark.

Complete step-by-step answer:
To calculate combination, we will have to calculate factorial
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)................4.3.2.1\]
To solve the given expression we can use the formula
\[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Given that,
\[n = 3,r = 2\]
Substituting the value of $n\,and\,r$ in the given formula
We get,
\[
  ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\
  ^3{C_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \\
  ^3{C_2} = \dfrac{{3!}}{{2!\left( 1 \right)!}} \\
    \\
 \]
Substitute the corresponding values in the below expression for 3!, 2!, and 1!, and simplify further .
\[
   = \dfrac{6}{{\left( 2 \right)\left( 1 \right)}} \\
   = 3 \\
  ^3{C_2} = 3 \\
 \]
Therefore 3 is the required solution that we have obtained after evaluating.
So, the correct answer is “Option 3”.

Note: A combination is a mathematical technique that determines the number of possible arrangements in a collection of its items where the order of the selection does not matters
Whereas a permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters. The distinction between permutation and combination is that in permutation, the order of the members is taken into account, while in combination, the order of the members is irrelevant.