Question
An ice cream store sells 25 flavors of ice cream. (a) How many different double-scoop cones (of 2 different flavors) can be made if order does not matter (for example, putting chocolate on top of vanilla is equivalent to putting vanilla on top of chocolate)? (b) How many different triple-scoop cones (of 3 different flavors) can be made if order does matter? (a) If order does not matter, then \( \square \) different double-scoop cones (of 2 different flavors) can be made. (Type a whole number.) (b) If order does matter, then \( \square \) different triple-scoop cones (of 3 different flavors) can be made. (Type a whole number.)
Ask by Love Garza. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
(a) 300 different double-scoop cones can be made.
(b) 13,800 different triple-scoop cones can be made.
Solution
1. For a double-scoop cone with 2 different flavors where order does not matter, we use combinations:
\[
\binom{25}{2} = \frac{25 \times 24}{2} = 300.
\]
2. For a triple-scoop cone with 3 different flavors where order does matter, we use permutations:
\[
25 \times 24 \times 23 = 13800.
\]
(a) \(300\)
(b) \(13800\)
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Simplify this solution Mind Expander
To solve the problems: (a) For double-scoop cones where order doesn't matter, we can use combinations. The formula for combinations is \( \binom{n}{r} \), where \( n \) is the total number of flavors and \( r \) is the number of flavors selected. So, we have: \[ \binom{25}{2} = \frac{25!}{2!(25-2)!} = \frac{25 \times 24}{2 \times 1} = 300 \] Thus, there are **300 different double-scoop cones**. (b) For triple-scoop cones where order does matter, we use permutations. The formula for permutations is \( P(n, r) = \frac{n!}{(n - r)!} \). Here, we find: \[ P(25, 3) = 25 \times 24 \times 23 = 13,800 \] Thus, there are **13,800 different triple-scoop cones**.
