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In this exercise, give an expression for the answer using permutation notation, combination notation, factorial notation, or other operations. Then evaluate. Baskin-Robbins Ice Cream. Burt Baskin and Irv Robbins began making ice cream in \(1945 .\) Initially they developed 31 flavors - one for each day of the month. (Source: Baskin-Robbins) a) How many 2 -dip cones are possible using the 31 original flavors if order of flavors is to be considered and no flavor is repeated? b) How many 2 -dip cones are possible if order is to be considered and a flavor can be repeated? c) How many 2 -dip cones are possible if order is not considered and no flavor is repeated?

Short Answer

Expert verified
a) 930 2-dip cones. b) 961 2-dip cones. c) 465 2-dip cones.

Step by step solution

01

Step 1a - Identifying the type of problem for part (a)

For part (a), we need to determine the number of ways to choose 2 different flavors out of the 31 original flavors if the order of flavors matters and no flavor is repeated. This is a permutation problem since order is important.
02

Step 2a - Using permutation notation for part (a)

Using permutation notation, the number of permutations of 2 out of 31 flavors is given by P(31, 2) = 31! / (31-2)! = 31! / 29!
03

Step 3a - Evaluating the permutation for part (a)

Evaluating the above expression: P(31, 2) = 31 * 30 = 930
04

Step 1b - Identifying the type of problem for part (b)

For part (b), we need to determine the number of ways to choose 2 flavors out of the 31 flavors if the order of flavors matters and a flavor can be repeated. This is a problem of ordered selection with repetition allowed.
05

Step 2b - Using multiplication for part (b)

There are 31 choices for the first flavor, and 31 choices for the second flavor since repetition is allowed. Therefore, the number of 2-dip cones is 31 * 31.
06

Step 3b - Evaluating the expression for part (b)

Evaluating the above expression: 31 * 31 = 961
07

Step 1c - Identifying the type of problem for part (c)

For part (c), we need to determine the number of ways to choose 2 different flavors out of the 31 flavors if the order of flavors does not matter and no flavor is repeated. This is a combination problem since order is not important.
08

Step 2c - Using combination notation for part (c)

Using combination notation, the number of combinations of 2 out of 31 flavors is given by C(31, 2) = 31! / [2!(31-2)!] = 31! / [2!*29!].
09

Step 3c - Evaluating the combination for part (c)

Evaluating the above expression: C(31, 2) = (31 * 30) / (2 * 1) = 465

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial Notation
Factorial notation is a mathematical notation that helps in counting and permutations. It is represented by an exclamation mark (!). For instance, the factorial of a number n, denoted as n!, is the product of all positive integers up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.
It's powerful for problems where order matters and rearrangements are required. In the context of the original exercise, it helps evaluate permutations and combinations.
Permutation Problems
Permutations are arrangements where order matters. When dealing with permutations, the number of ways to arrange a subset of items from a larger set is calculated. In the exercise, part (a) asks for the number of 2-dip cones with 31 flavors where order matters and no repetition. This is represented by the permutation formula \( P(n, r) = \frac {n!} {(n-r)!}\), where n is the total items, and r is the subset. Here, it's \( P(31, 2) = \frac {31!} {29!}\), which simplifies to 31 * 30 = 930.
Combination Problems
Combinations are selections where the order does not matter. When calculating combinations, the number of ways to choose a subset of items from a larger set, regardless of order, is determined.
Part (c) of the exercise is a combination problem. The formula used is \( C(n, r) = \frac {n!} {r!(n-r)!} \). For 31 flavors, choosing 2 where order does not matter, it's \( C(31, 2) = \frac {31!} {2! 29!} \), which simplifies to \( \frac{31 × 30}{2 × 1} \). This equals 465.
This concept is used in various areas, such as probability, where order is irrelevant.

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Most popular questions from this chapter

Give your answer using permutation notation, factorial notation, or other operations. Then evaluate. How many permutations are there of the letters in each of the following words, if all the letters are used without repetition? How many permutations are there of the letters of the word EDUCATION if the letters are taken 4 at a time?

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Give your answer using permutation notation, factorial notation, or other operations. Then evaluate. How many permutations are there of the letters in each of the following words, if all the letters are used without repetition? How many code symbols can be formed using 5 of the 6 letters \(A, B, C, D, E, F\) if the letters: a) are not repeated? b) can be repeated? c) are not repeated but must begin with D? d) are not repeated but must begin with DE?

Expand and simplify: \(\frac{(x+h)^{13}-x^{13}}{h}\)

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