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In Exercises 51 - 56, evaluate \( _nC_r \) using the formula from this section. \( _{25}C_0 \)

Short Answer

Expert verified
The value of \( _{25}C_0 \) is 1.

Step by step solution

01

Identify the values

We are given the following to evaluate: \( _{25}C_0 \). So in this case, \(n = 25\) and \(r = 0\).
02

Apply the Combination Formula

The formula for the combination is \(_nC_r = \frac{n!}{r!(n-r)!}\). When you put the values in the formula, we get \(_{25}C_0 = \frac{25!}{0!(25-0)!}\).
03

Simplify the expression

0! is always equal to 1. Replacing 0! with 1 in our formula gives us \(_{25}C_0 = \frac{25!}{1 * 25!} = 1\) as both 25! in the numerator and the denominator will cancel out each other.

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Key Concepts

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

Factorial
A factorial, denoted by the exclamation mark (!), is a basic mathematical function used to multiply a series of descending natural numbers. For any non-negative integer \( n \), the factorial is defined as follows:
  • \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \)
This means that you start at the number \( n\) and multiply it by each natural number before it until you reach 1.

For example:
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \( 3! = 3 \times 2 \times 1 = 6 \)
  • \( 1! = 1 \)
  • By convention, \( 0! = 1 \)
The factorial function is essential in combinations, permutations, and other areas of mathematics because it simplifies the calculation of large numbers. The concept of a factorial helps reduce complex multiplication into simple, manageable steps, making calculations quicker and more efficient.
Binomial Coefficient
The binomial coefficient, often represented as \( \binom{n}{r} \) or \( _nC_r \), is a fundamental part of combinatorics that describes the number of ways to choose \( r \) items from \( n \) total items without regard to the order of selection.

The formula for the binomial coefficient is:
  • \( _nC_r = \frac{n!}{r!(n-r)!} \)
This formula can be broken down into the following parts:
  • The numerator, \( n! \), accounts for arranging all \( n \) items.
  • The denominator, \( r! \), adjusts for the permutations of the \( r \) selected items.
  • The additional factor \( (n-r)! \) ensures that we do not overcount the permutations of the \( n-r \) remaining items.
Let's consider the solved example from the original exercise: we are asked to calculate \( _{25}C_0 \).

Using the above formula, we get:
  • \( _{25}C_0 = \frac{25!}{0!(25-0)!} = \frac{25!}{1 \times 25!} = 1 \)
Here, the factorial terms cancel each other out, simplifying the calculation significantly. The result is exactly as expected: when choosing zero items from a set, there is only one possible way, which is to choose nothing at all.
Permutation and Combination
Permutations and combinations, though related, describe different ways to arrange or select objects.
  • Permutation: Permutations consider the order of arrangement. If the sequence or positioning of items matters, you're dealing with permutations. For example, arranging books on a shelf in a specific sequence is a permutation.
  • Mathematically, it is expressed as \( _nP_r = \frac{n!}{(n-r)!} \) where you arrange \( r \) items picked from \( n \) distinct items.
  • Combinations, on the other hand, ignore the order of items. Only the selection of items matters, not how they are arranged. This concept is captured in the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) which we've discussed in detail previously.
An easy way to remember this is:
  • If order doesn't matter, use combinations.
  • If order does matter, use permutations.
For example, in a lottery draw, the order in which the numbers are selected doesn't matter, hence a combination is used. However, if you're organizing a race, the order in which people finish matters, so you would use permutations.
Understanding permutations and combinations is crucial in problem-solving because they provide a framework to tackle a wide range of real-world scenarios, from probability calculations to decision-making processes.

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Most popular questions from this chapter

In Exercises 25 - 30, find the probability for the experiment of tossing a six-sided die twice. The sum is odd and no more than \( 7 \).

In Exercises 35 - 38, you are given the probability that an event will happen. Find the probability that the event will not happen. \( P(E) = 0.36 \)

Consider a group of people. (a) Explain why the following pattern gives the probabilities that the people have distinct birthdays. \( n = 2: \dfrac{365}{365} \cdot \dfrac{364}{365} = \dfrac{365 \cdot 364}{365^2} \) \( n = 3: \dfrac{365}{365} \cdot \dfrac{364}{365} \cdot \dfrac{363}{365} = \dfrac{365 \cdot 364 \cdot 363}{365^3} \) (b) Use the pattern in part (a) to write an expression for the probability that \( n = 4 \) people have distinct birthdays. (c) Let \( P_n \) be the probability that the \( n \) people have distinct birthdays. Verify that this probability can be obtained recursively by \( P_1 = 1 \) and \( P_n = \dfrac{365 - (n - 1)}{365} P_{n - 1} \). (d) Explain why \( Q_n = 1 - P_n \) gives the probability that at least two people in a group of \( n \) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \( \dfrac{1}{2} \)? Explain.

You are dealt five cards from an ordinary deck of \( 52 \) playing cards. In how many ways can you get (a) a full house and (b) a five-card combination containing two jacks and three aces? (A full house consists of three of one kind and two of another. For example, \( A-A-A-5-5 and K-K-K-10-10 \) are full houses.)

In Exercises 21 - 24, find the probability for the experiment of selecting one card from a standard deck of \( 52 \) playing cards. The card is a red face card.

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