Holly chooses seven out of nine plants for the front of her house with the remaining plants to be used on the side of the house.

If order does not matter, in how many ways can she choose the plants for the front of the house?

7
9
21
36

1 answer

To find the number of ways Holly can choose 7 plants out of 9, we can use the concept of combinations. The formula for combinations is given by:

\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]

where \(n\) is the total number of items to choose from, \(r\) is the number of items to choose, and "!" denotes factorial (the product of all positive integers up to that number).

In this case, \(n = 9\) and \(r = 7\). The number of ways Holly can choose 7 plants from 9 is:

\[ \binom{9}{7} = \frac{9!}{7!(9 - 7)!} = \frac{9!}{7! \cdot 2!} \]

Calculating this step by step:

  1. Calculate the factorials:

    • \(9! = 9 \times 8 \times 7!\)
    • \(7! = 7!\)
    • \(2! = 2 \times 1 = 2\)
  2. Substitute into the combination formula:

\[ \binom{9}{7} = \frac{9 \times 8 \times 7!}{7! \cdot 2} = \frac{9 \times 8}{2} \]

  1. Simplify:

\[ \frac{9 \times 8}{2} = \frac{72}{2} = 36 \]

Thus, the number of ways Holly can choose 7 plants for the front of her house is \( \boxed{36} \).