Question

In a class there are 20 boys and 15 girls. In how many ways can 2 boys and 2 girls be selected? (1) \({ }^{35} \mathrm{C}\) (2) \({ }^{35} \mathrm{C}\), (3) \({ }^{20} \mathrm{C}_{2} \times{ }^{15} \mathrm{C}_{2}\) (4) \(20 \times 15\)

Short answer

Answer: There are 19,950 ways to select 2 boys and 2 girls from the class.

Step-by-step solution

Calculate combinations for boys

First, calculate the number of ways to select 2 boys from a group of 20 boys. We can use the combination formula: \(C(n,r) = \frac{n!}{r!(n-r)!}\). In this case, n=20 and r=2, so we have: \(C(20,2) = \frac{20!}{2!(20-2)!} = \frac{20!}{2!18!} = \frac{20\times19}{2\times1} = 10\times19 = 190\). So there are 190 ways to select 2 boys from a group of 20 boys.

Calculate combinations for girls

Next, calculate the number of ways to select 2 girls from a group of 15 girls. Again using the combination formula with n=15 and r=2: \(C(15,2) = \frac{15!}{2!(15-2)!} = \frac{15!}{2!13!} = \frac{15\times14}{2\times1} = 7\times15 = 105\). So there are 105 ways to select 2 girls from a group of 15 girls.

Combine boys and girls

Now, to find the total number of ways to select 2 boys and 2 girls, we will multiply the number of ways to select the boys (190) by the number of ways to select the girls (105): Total ways = \(190 \times 105 = 19950\). The correct answer is: (3) \({ }^{20} \mathrm{C}_{2} \times{ }^{15} \mathrm{C}_{2}= 19950\).

Key concepts

Combinatorics

Combinatorics is a branch of mathematics focused on counting, enumeration, and the study of combinatory structures and their properties. One of the most common problems in combinatorics involves finding the number of ways a certain event can occur given a set of items or choices. It includes various subfields such as graph theory, enumeration, and design theory, but when we deal with questions like selecting students from a class, we are dealing with combinatorial enumeration.

For example, let's imagine we want to form a committee from a group of people, or we wish to organize a sequence of races with different participants every time. By using combinatorics, we can systematically count these possibilities without having to list each one, which becomes especially useful when dealing with large sets. The basis of solving such problems typically involves understanding the fundamental principles of counting, such as the Rule of Product, the Rule of Sum, permutations, and combinations, which are each suited to different types of combinatorial situations.

Factorial Notation

Factorial notation is a concise way to represent the product of an integer and all the positive integers below it down to 1, and it's signified by the exclamation mark (!). For any positive integer n, the factorial of n, represented as n!, is calculated as:
\[ n! = n \times (n - 1) \times (n - 2) \times \ldots \times 3 \times 2 \times 1 \]
For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Zero factorial, \( 0! \), is defined to be 1. Factorials are fundamental in combinatorics for calculating permutations and combinations because these concepts often involve multiplying a sequence of descending natural numbers. They start to increase rapidly with each additional number; consequently, factorials are notably used in situations where the arrangement or selection order is significant, such as in permutations.

Permutations and Combinations

Permutations and combinations are two concepts that are the cornerstone of combinatorics and are used for determining the number of different ways in which elements from a set can be arranged or selected.

• Permutations: This relates to the arrangement of objects in a specific order. The number of permutations of n distinct objects taken r at a time is represented as \( P(n, r) \) and calculated using the formula: \[ P(n, r) = \frac{n!}{(n-r)!} \]
• Combinations: Unlike permutations, combinations refer to the selection of objects where the order does not matter. The number of combinations of n distinct objects taken r at a time, denoted as \( C(n, r) \) or \( ^nC_r \), is given by the formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]

From the original exercise, selecting 2 boys out of 20 and 2 girls out of 15, we are dealing with combinations since the order in which the boys or girls are selected is not relevant. By applying the combination formula to each gender separately and then multiplying the two results, we get the total number of ways to select the students. Through understanding and applying these fundamental concepts of permutations and combinations, we can solve a wide array of complex combinatorial problems in mathematics.