Problem 30 - Calculus for Biology and Medicine

Question

Twelve children are divided up into three groups, of five, four, and three children, respectively. In how many ways can this be done if the order within each group is not important?

Step-by-Step Solution

Answer

There are 27,720 ways to divide the 12 children into groups of 5, 4, and 3.

Step 1: Understand the problem

We need to divide 12 children into three groups with sizes 5, 4, and 3. The order within each group doesn't matter. We will look into different combinations that respect these group sizes.

Step 2: Choose children for the first group

Select 5 children out of 12 to form the first group. This can be done in \( \binom{12}{5} \) ways, where \( \binom{n}{k} \) denotes the binomial coefficient for choosing \( k \) items out of \( n \).

Step 3: Choose children for the second group

After forming the first group, 7 children remain. Select 4 children out of these 7 for the second group. This can be done in \( \binom{7}{4} \) ways.

Step 4: Assign children to the third group

The remaining 3 children automatically form the third group since there are no remaining choices to be made. This step can be done in exactly 1 way, since \( \binom{3}{3} \) is 1.

Step 5: Calculate the total number of ways

Multiply the number of ways to form each group: \( \binom{12}{5} \binom{7}{4} \binom{3}{3} \). This gives us the total number of ways to divide the 12 children into groups of 5, 4, and 3.

Step 6: Solve the binomial coefficients

Calculate each binomial coefficient:- \( \binom{12}{5} = \frac{12!}{5!(12-5)!} = 792 \)- \( \binom{7}{4} = \frac{7!}{4!(7-4)!} = 35 \)- \( \binom{3}{3} = 1 \)

Step 7: Compute the final answer

Multiply the values obtained in Step 6: \[ 792 \times 35 \times 1 = 27,720 \]Therefore, there are 27,720 different ways to divide these 12 children into groups of 5, 4, and 3.

Key Concepts

Binomial Coefficient

The binomial coefficient is a useful mathematical concept used to compute the number of ways to choose a subset of items from a larger set, where the order of selection does not matter. It's expressed as \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items we want to select. The formula is given by

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Factorial \( n! \) represents the product of all whole numbers from 1 to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
In the context of our problem, using binomial coefficients allows us to determine the number of ways to pick groups of children from the set of 12.
This tool is essential in solving problems where subgroup selection from a set is necessary, like dividing people into teams or selecting committee members.

Grouping Problems

Grouping problems involve dividing a set of items or people into specific groups or categories. In our exercise, the task is to divide 12 children into groups of 5, 4, and 3.
The key to solving grouping problems is to look at the selection in parts: first, form the largest group, then the next, and the smallest one last.

Here’s a simple strategy:

• First, choose the number of items for the first group. Example: select 5 children out of 12.
• Then, choose items for the next group from the reduced set. Example: select 4 out of the remaining 7 children.
• The smallest group is automatically formed, needing no further choice.

Such systematic approach ensures no item is left unaccounted, and all group size constraints are met correctly. Grouping problems are common in combinatorial scenarios where each group has a definitive size.

Permutations and Combinations

Understanding permutations and combinations is crucial in solving problems involving multiple selections and arrangements.

Here’s a brief look:

• Permutations deal with arrangements where order matters. For example, arranging books on a shelf.
• Combinations focus on selection where order doesn’t matter, such as choosing a team from a larger group.

In our specific exercise, combinations are used since the order of children within the group does not contribute to additional distinct outcomes.
For instance, selecting three children from a group of five is a combination problem because it doesn’t matter which child comes first.
This principal distinction simplifies the counting task significantly in many practical scenarios, allowing us to focus only on the number chosen rather than their order.