Samantha’s parents took her to a theme park and told her that she could pick out three toys from the gift shop if there are nine different toys in total how many different selections of three toys could she make write our Pascal’s triangle from row 9 and use it to solve the problem

1 answer

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.