To determine how many different selections of three toys Samantha can make from a total of nine different toys, we can use combinations, which can be represented mathematically as \( C(n, r) \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
In this case, we need to calculate \( C(9, 3) \):
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
Substituting the values, we have:
\[ C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \]
Now, we can simplify this:
\[ 9! = 9 \times 8 \times 7 \times 6! \] So,
\[ C(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{3! \times 6!} = \frac{9 \times 8 \times 7}{3!} \]
Calculating \( 3! \):
\[ 3! = 3 \times 2 \times 1 = 6 \]
Now substituting back, we get:
\[ C(9, 3) = \frac{9 \times 8 \times 7}{6} = \frac{504}{6} = 84 \]
So, there are 84 different ways Samantha can select 3 toys from 9 different toys.
Using Pascal’s Triangle
To also illustrate this using Pascal's Triangle, we recognize that the entry at row \( n \) and column \( r \) corresponds to \( C(n, r) \). The rows of Pascal's Triangle begin with row 0 at the top:
- Row 0: \( 1 \)
- Row 1: \( 1, 1 \)
- Row 2: \( 1, 2, 1 \)
- Row 3: \( 1, 3, 3, 1 \)
- Row 4: \( 1, 4, 6, 4, 1 \)
- Row 5: \( 1, 5, 10, 10, 5, 1 \)
- Row 6: \( 1, 6, 15, 20, 15, 6, 1 \)
- Row 7: \( 1, 7, 21, 35, 21, 7, 1 \)
- Row 8: \( 1, 8, 28, 56, 28, 8, 1 \)
- Row 9: \( 1, 9, 36, 84, 36, 9, 1 \)
Here, the entry at row 9, column 3 (which counts as \( C(9, 3) \)) is 84.
Thus, we've confirmed our earlier calculation: Samantha can select her toys in 84 different ways.