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One quarter of the five-element subsets of \(\\{1,2,3, \ldots, n\\}\) contain the element 7 . Determine \(n(\geq 5)\)

Short Answer

Expert verified
The original set must have 21 elements for exactly one quarter of its five-element subsets to contain the number 7.

Step by step solution

01

Understand Counting and Subsets

Before we start solving the problem, it is important to understand how the subsets are formed and counted. In this case, we are creating sets of five elements from the set \{1,2,3, ..., n\}. These sets are formed without regards to the order of elements, hence they are combinations.
02

Setup the equation

Given that one quarter of the subsets including 7, this means the number of five-element subsets with 7 would equal to one quarter of the total number of five-element subsets. Therefore, we can set an equation to represent this: \[ \binom{n-1}{4} = \frac{1}{4} \binom{n}{5}\] The \(\binom{n-1}{4}\) represents the combinations of choosing 4 remaining elements from n-1 elements (since one element, 7, is already chosen), and \(\binom{n}{5}\) represents the total combinations of choosing 5 elements from n elements.
03

Solve the equation

To solve this equation, simplify it first using the formula for combinations: \( \frac{(n-1)!}{4!(n-1-4)!}=\frac{1}{4} \frac{n!}{5!(n-5)!}\). Simplify further: \(\frac{(n-1)!}{4!(n-5)!} = \frac{n!}{5!(n-5)!}\). It simplifies further into: \( n-1 = 4n/5\). Solving for n, we get \(n = 21\).
04

Interpret the result

The result means that in order for one quarter of the five-element subsets to contain the element 7, there must be 21 elements in the original set.

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Key Concepts

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

Counting and Subsets
To delve into the realm of combinatorics, a foundational concept is counting and determining subsets within a set. A subset is a portion, or a smaller set, taken from a larger set of elements. The process of identifying all possible subsets - especially those of a particular size - requires an understanding of combinations. In our exercise, we are tasked to find subsets of size five from a larger set beginning with 1 and ending with an undetermined number, n.

Consider you have a set of books on a shelf, and wish to pick five to take on a holiday. It doesn't matter in which order you pick them; what matters is which books you choose. The total number of different selections (subsets) of five books forms a combination from your entire collection. This principle governs the problem we're exploring: taking five elements from a set that includes numbers from 1 to n.
Combinations
The concept of combinations relates closely to the idea of counting and subsets. When dealing with combinations, the order of selection is not important. It is the mathematical method we use to count the number of ways in which we can select a particular group of items from a larger pool without considering the sequence of selection.

For instance, when choosing ice cream flavors, if you want to select three out of ten flavors, it doesn't matter whether you pick vanilla, chocolate, and strawberry, or strawberry, chocolate, and vanilla -- the resultant combination is the same. Mathematically, combinations are denoted using binomial coefficients, which are read as 'n choose k' and written as \(\binom{n}{k}\)\. The formula for a combination is given by: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n!\) is the factorial of n, which is the product of all positive integers up to n.
Solving Combinatorial Equations
When it comes to solving combinatorial equations, we are often required to set up an equation based on the problem's conditions and solve for an unknown. These kinds of problems can frequently be represented by equations involving binomial coefficients. As observed in our example, we're given that exactly one quarter of the five-element subsets contain a specific number, 7, thereby enabling us to form an equation expressing that relationship.

With this in hand, our next step is to apply mathematical methods (like simplification of factorial expressions) to find the solution to our problem. It's like solving a puzzle: combining pieces of information until the bigger picture becomes clear. Here, we reduce the combinatorial equation step by step until we isolate the variable n, providing us with the precise size of our set.

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Most popular questions from this chapter

For \(\ddot{u}=\mathbf{Z}^{+}\)let \(A \subseteq \vartheta\) where \(A=\\{1,2,3,4,5,7,8,10,11,14,17,18\\}\). a) How many subsets of \(A\) contain six elements? b) How many six-element subsets of \(A\) contain four even integers and two odd integers? c) How many subsets of \(A\) contain only odd integers? d) How many of the subsets in part (c) contain the integers 3 and 7 ?

a) In how many ways can a teacher divide a group of seven students into two teams each containing at least one student? two students? b) Answer part (a) upon replacing seven with a positive integer \(n \geq 4\).

For \(A=\\{1,2,3,4,5,6,7\\}\), determine the number of a) subsets of \(A\). b) nonempty subsets of \(A\). c) proper subsets of \(A\). d) nonempty proper subsets of \(A\). e) subsets of \(A\) containing three elements. f) subsets of \(A\) containing 1,2 . g) subsets of \(A\) containing five elements, including \(1,2 .\) h) proper subsets of \(A\) containing 1,2 . i) subsets of \(A\) with an even number of elements. j) subsets of \(A\) with an odd number of elements. k) subsets of \(A\) with an odd number of elements, including the element \(3 .\)

Professor Diane gave her chemistry class a test consisting of three questions. There are 21 students in her dass, and every student answered at least one question. Five students did not answer the first question, seven failed to answer the second question, and six did not answer the third question. If nine students answered all three questions, how many answered exactly one question?

a) In chess, the king can move one position in any direction. Assuming that the king is moved only in a forward manner (one position up, to the right, or diagonally northeast), along how many different paths can a king be moved from the lower-left corner position to the upper-right corner position on the standard \(8 \times 8\) chessboard? b) For the paths in part (a), what is the probability that a path contains (i) exactly two diagonal moves? (ii) exactly two diagonal moves that are consecutive? (iii) an even number of diagonal moves?

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