- The combinations are calculated using the formula \({}_n C_k = \frac{n!}{k!(n-k)!}\).
- \({}_3 C_2 = \boxed{3}\)
- \({}_3 C_3 = \boxed{1}\)
- \({}_4 C_2 = \boxed{6}\)
Explanation
- Understand the Combination Formula
We are tasked with calculating specific values for combinations using the standard definition of the combination formula, which is given by:
\[ {}_n C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \(n!\) denotes the factorial of \(n\), defined as \(n! = n \times (n-1) \times \dots \times 1\), and \(0! = 1\). Combinations represent the number of ways to select a subset of \(k\) items from a larger set of \(n\) distinct items, where the order of selection does not matter.
- Calculate \({}_3 C_2\)
To find \({}_3 C_2\), we substitute \(n=3\) and \(k=2\) into the formula:
\[ {}_3 C_2 = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} \]
Calculating the factorials:
\(3! = 3 \times 2 \times 1 = 6\)
\(2! = 2 \times 1 = 2\)
\(1! = 1\)
Therefore:
\[ {}_3 C_2 = \frac{6}{2 \times 1} = \frac{6}{2} = 3 \]
So, \({}_3 C_2 = 3\).
- Calculate \({}_3 C_3\)
To find \({}_3 C_3\), we substitute \(n=3\) and \(k=3\) into the formula:
\[ {}_3 C_3 = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} \]
Recalling that \(0! = 1\):
\[ {}_3 C_3 = \frac{6}{6 \times 1} = \frac{6}{6} = 1 \]
So, \({}_3 C_3 = 1\). This makes sense intuitively: there is only one way to choose all 3 items from a set of 3.
- Calculate \({}_4 C_2\)
To find \({}_4 C_2\), we substitute \(n=4\) and \(k=2\) into the formula:
\[ {}_4 C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} \]
Calculating the factorials:
\(4! = 4 \times 3 \times 2 \times 1 = 24\)
\(2! = 2\)
So:
\[ {}_4 C_2 = \frac{24}{2 \times 2} = \frac{24}{4} = 6 \]
This result matches the entry found in Pascal's triangle, where row 4 is \(1, 4, 6, 4, 1\), and the second index (starting from 0) is 6. Thus, \({}_4 C_2 = 6\).
Examples
Combinations are used in everyday life whenever you need to determine the number of ways to pick a group from a larger set. For example, if you have 4 different colors of paint and want to choose 2 to mix for a new shade, the number of possible unique combinations is \({}_4 C_2 = \frac{4!}{2!2!} = 6\). Understanding this helps in everything from planning lottery odds to organizing tournament matchups!