Question

Let \(S=\\{1,2,3, \ldots, 9\\}\). For \(k=1,2, \ldots, 5\), let \(N_{k}\) be the number of subsets of \(S\), each containing five elements out of which exactly \(k\) are odd. Then \(N_{1}+N_{2}+N_{3}+N_{4}+N_{5}=\) [A] 210 [B] 252 [C] 125 [D] 126

Short answer

126

Step-by-step solution

Identify the Number of Odd and Even Elements

First, count the number of odd and even numbers in the set S. The set S = {1, 2, 3, ..., 9} contains 5 odd numbers (1, 3, 5, 7, 9) and 4 even numbers (2, 4, 6, 8).

Calculate Individual N_k Values

Use combinations to find the number of ways to select k odd numbers and (5-k) even numbers for each N_k. For N_1, choose 1 odd and 4 even: \( \binom{5}{1}\binom{4}{4} \).For N_2, choose 2 odd and 3 even: \( \binom{5}{2}\binom{4}{3} \).For N_3, choose 3 odd and 2 even: \( \binom{5}{3}\binom{4}{2} \).For N_4, choose 4 odd and 1 even: \( \binom{5}{4}\binom{4}{1} \).For N_5, choose 5 odd and 0 even: \( \binom{5}{5}\binom{4}{0} \).

Compute the Combinations

Evaluate each combination from the previous step:\( N_1 = \binom{5}{1}\binom{4}{4} = 5 \times 1 = 5 \).\( N_2 = \binom{5}{2}\binom{4}{3} = 10 \times 4 = 40 \).\( N_3 = \binom{5}{3}\binom{4}{2} = 10 \times 6 = 60 \).\( N_4 = \binom{5}{4}\binom{4}{1} = 5 \times 4 = 20 \).\( N_5 = \binom{5}{5}\binom{4}{0} = 1 \times 1 = 1 \).

Add the Values of N_k

Now add the calculated values of N_k to find the total number of subsets: \( N_1 + N_2 + N_3 + N_4 + N_5 = 5 + 40 + 60 + 20 + 1 = 126 \).

Select the Correct Answer

Based on the sum, the correct answer is: 126. Thus, the correct option is [D] 126.

Key concepts

Combination and Permutation

Understanding the difference between combinations and permutations is essential for solving problems related to subset selection. Let's simplify these concepts:

In mathematics, a permutation is an arrangement of items in a specific order. For example, if you were to list your top three favorite fruits, the order in which you list them matters. So, 'apple, banana, cherry' is considered a different permutation than 'banana, cherry, apple'.

On the other hand, a combination is a selection of items where order doesn't matter. Using the same fruit example, selecting 'apple, banana, cherry' as your favorite fruits is the same combination as 'cherry, apple, banana'.

In the exercise given, we are focused on combinations because selecting subsets of numbers from set S doesn’t depend on the order of the elements. It’s important to grasp that when the problem involves selecting items and the order isn't factored in, you're working with combinations, denoted as \( \binom{n}{k} \), which calculates the number of ways to choose k items from a larger set of n items.

Subset Selection

Subset selection is the process of choosing a certain number of elements from a larger set, where each element has an equal chance of being chosen. This concept is widely used in the field of statistics, probability, and combinatorics. For the problem at hand, we're interested in selecting subsets of five elements from set S where a certain number of elements is odd.

The key strategy in subset selection is deciding which elements must be in the subset and which could possibly be included. In our exercise, we must select exactly k odd numbers for each subset, and the rest will be even numbers. This step-by-step process makes it manageable to apply the formulas for combinations. Improving understanding of this process can be done by practicing different selections and manually counting the subsets before confirming with the combination formulas.

Counting Principles in Math

Counting principles, such as the fundamental counting principle, permutations, and combinations, form the basis of enumerative combinatorics. They provide systematic ways to quantify the number of possible configurations in a set or to determine the size of various sets.

The exercise provided is a practical application of the addition principle, which states that if you have several groups of items, you can find the total number of items by adding the number of items in each group. This principle aids in reaching the solution when you break down the problem into smaller, manageable parts, i.e., calculating \(N_k\) for each value of k and then summing them all up.

Understanding how to use counting principles effectively can significantly improve problem-solving skills in mathematics. It's useful to remember these principles when faced with a task that requires you to count or enumerate possible arrangements or selections.