/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Evaluate. $$C(9,4)$$... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate. $$C(9,4)$$

Short Answer

Expert verified
The value of \(C(9,4)\) is 126

Step by step solution

01

Identify n and k

In the combination notation \(C(n,k)\), 'n' represents the total number of elements, and 'k' represents the number of elements to select. Here \(C(9,4)\), so n = 9 and k = 4.
02

Apply the combination formula

The combination can be found using the formula \[C(n,k) = \frac{n!}{k!(n-k)!}\]. Therefore, for this problem we get: \[C(9,4) = \frac{9!}{4!(9-4)!}\].
03

Calculate factorials

Now we have to calculate the factorials. Recall that the factorial of a number is the product of every whole number from that number down to 1. Thus, we have: \[9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880\] \[4! = 4 × 3 × 2 × 1 = 24\] \[(9-4)! = 5! = 5 × 4 × 3 × 2 × 1 = 120 \]
04

Substitute factorials into the formula

When we substitute our results back into the formula, the calculation becomes: \[C(9,4) = \frac{362880}{24 × 120}\]
05

Perform the division

The final step is the arithmetic—carry out the division to find the answer: \[C(9,4) = \frac{362880}{2880} = 126\]

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Key Concepts

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

Factorial
The concept of factorial is a crucial part of understanding combinations and permutations. When we talk about the factorial of a number, we mean the product of all positive integers from that number down to 1. For instance, the factorial of 5, denoted as \(5!\), is calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\). This gives us the total number of ways to arrange 5 distinct items.

Factorials are handy when dealing with problems that may involve arranging or selecting items. They are a building block for formulas used in advanced mathematical topics such as algebra and calculus.

When you see \(n!\) in the context of combinations, it typically serves to quantify the numerous ways to arrange or select a group of items. In our example, \(9!\) would signify every possible permutation of 9 items, which equals 362880.
Combination Formula
The combination formula is essential in finding how many ways we can select items from a larger set, disregarding the order of selection. The formula for a combination, \(C(n,k)\), or \(\binom{n}{k}\), is expressed mathematically as:
  • \[ C(n,k) = \frac{n!}{k!(n-k)!} \]
In this formula:
  • \(n\) represents the total number of items available.
  • \(k\) is the number of items to be chosen.

The formula helps determine the number of distinct groups that can be formed from a set without considering the order of elements within each group. By dividing by \(k!\), it accounts for the redundancies that occur when arranging \(k\) items, yielding non-repeating combinations.

In our example, \(C(9,4)\), this formula was used to find how many different ways we can form a group of 4 people from a total of 9.
Binomial Coefficient
The binomial coefficient is another term for combination and is often denoted as \(\binom{n}{k}\). It plays a significant role in combinatorics and statistical calculations, especially within binomial expansions and probability theory.

In mathematics, the binomial coefficient \(\binom{n}{k}\) describes how many different combinations can be made of \(k\) elements chosen from a set of \(n\) elements, disregarding order. This concept provides a convenient way to calculate the number of subsets of a certain size that can be selected from a larger set.

You'll frequently encounter binomial coefficients in problems involving probabilities, such as calculating the likelihood of a certain number of successes in a series of independent experiments. The efficient calculation of these coefficients often relies on the understanding and application of the factorial concept and combination formula as explored in our example with \(C(9,4)\), where we found there are 126 unique groups of 4 that can be formed from 9 items.

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Most popular questions from this chapter

In this set of exercises, you will use sequences to study real-world problems. Knitting Knitting, whether by hand or by machine, uses a sequence of stitches and proceeds row by row. Suppose you knit 100 stitches for the bottommost row and increase the number of stitches in each row thereafter by 4 This is a standard way to make the sleeve portion of a sweater. (a) What type of sequence does the number of stitches in each row produce: arithmetic, geometric, or neither? (b) Find a rule that gives the number of stitches in the nth row. (c) How many rows must be knitted to end with a row of 168 stitches?

Concepts This set of exercises will draw on the ideas presented in this section and your general math background. \- A sequence \(b_{0}, b_{1}, b_{2}, \ldots\) has the property that \(b_{n}=\) \(\left(\frac{n+3}{n+2}\right) b_{n-1}\) for \(n=1,2,3, \ldots,\) where \(c\) is a positive constant to be determined. Find \(c\) if \(b_{2}=25\) and \(b_{4}=315\)

A wedding photographer lines up four people plus the bride and groom for a photograph. If the bride and groom stand side-by-side, how many different photographs are possible?

During the play of a card game, you have seen 20 of the 52 cards in the deck and none of them is the 4 of clubs. You need the 4 of clubs to win the game. What is the probability that you will win the game on the next card drawn?

Consider the following experiment: pick one coin out of a bag that contains one quarter, one dime, one nickel, and one penny. What is the probability of picking a quarter or a penny?

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