Question

Two numbers are selected at once from the set of integers 1 to 20 . Find the probability that the product of the numbers will be 24 . (1) \(\frac{3}{190}\) (2) \(\frac{3}{380}\) (3) \(\frac{4}{95}\) (4) \(\frac{3}{95}\)

Short answer

Answer: The probability is \(\frac{3}{190}\).

Step-by-step solution

List all the factor pairs of 24

We need to find all the factor pairs whose product is equal to 24. Since we are only considering integers between 1 and 20, we want to list the factor pairs within that range. Factor pairs of 24 are: 1 × 24 2 × 12 3 × 8 4 × 6 Notice that 1 × 24 is not in the range between 1 and 20, so we only have 3 valid factor pairs: (2, 12), (3, 8), and (4, 6).

Find the total possible pair combinations

There are 20 integers in the given range, and we are selecting 2 numbers without replacement. The number of ways to select 2 numbers from a set of 20 can be calculated using the combinations formula, which is denoted as C(n, k): C(n, k) = \(\frac{n!}{k!(n-k)!}\) For our problem, n = 20, and k = 2: C(20, 2) = \(\frac{20!}{2!(20-2)!}\) = \(\frac{20!}{2!18!}\) = \(\frac{20 \times 19}{2} = 190\) We have 190 possible pair combinations from the integers 1 to 20.

Calculate the probability

To find the probability that the product of the numbers will be 24, we divide the number of desired factor pairs by the total possible pair combinations as mentioned in step 2: Probability = \(\frac{\text{Number of desired pairs}}{\text{Total possible pairs}}\) There are 3 desired pairs (2, 12), (3, 8) and (4, 6) from step 1, and 190 total possible pair combinations from step 2. Probability = \(\frac{3}{190}\) So the answer is (1) \(\frac{3}{190}\).

Key concepts

Factor Pairs

When we talk about factor pairs, we're referring to two numbers that, when multiplied together, give us a product. In the context of our exercise, the product is the number 24. It's essential to list down all possible factor pairs that give us 24, because this will be pivotal in calculating the probability of selecting such pairs from a set.

In the provided solution, we identified three valid factor pairs within the range of integers from 1 to 20: (2, 12), (3, 8), and (4, 6). It's important to note that when working with factor pairs, only unique sets of numbers are considered; the order in which we multiply the factors doesn't matter, (2 x 12) is the same as (12 x 2).

To assist students further, remember that factor pairs can be found by dividing the product by each of its factors. For instance, when we divided 24 by 1, 2, 3, and 4, we received 24, 12, 8, and 6, respectively, producing our factor pairs. Identifying factor pairs is a foundational skill in not just probability but also in topics such as number theory and algebra.

Combinations Formula

The combinations formula is instrumental when calculating the number of ways in which a set number of items can be selected from a larger pool where the order does not matter. In our example, we use this formula to determine how many different pairs can be formed from the numbers 1 to 20.

The generic formula for combinations is expressed as follows:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where \( n \) is the total number of items to choose from, \( k \) is the number of items to pick, and \( ! \) denotes the factorial operation, which is the product of an integer and all the integers below it down to 1.

Using this in our problem where \( n = 20 \) and \( k = 2 \), we find there are 190 possible combinations. To ensure students grasp this concept, emphasize that the combinations formula is used in scenarios where the sequence of selection is irrelevant, such as selecting a pair of numbers or a committee from a group.

Permutations and Combinations

The concepts of permutations and combinations are central to understanding arrangements in probability and combinatorics. Permutations consider the arrangement of items where the order does matter, unlike combinations where the order is irrelevant.

In permutations, which are noted as \( P(n, k) \), each arrangement is unique based on the sequence of items. On the other hand, as we saw with the combinations formula, combinations are more about selection without regard to the sequencing of items.

To draw a clear distinction for students, imagine selecting a first and second place in a race (permutation) versus selecting two members to be on a committee (combination). With the race, the order in which the runners finish matters, but for the committee, it does not.

To relate this back to our exercise, we used the concept of combinations since the product of two numbers (e.g., 2 and 3 to get 6) is the same regardless of the order in which they are multiplied. Therefore, for probability problems that involve products or unordered selections, combinations are the go-to calculation method.