Chapter 9: Problem 55
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
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:
For example:
- \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \)
For example:
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 1! = 1 \)
- By convention, \( 0! = 1 \)
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:
Using the above formula, we get:
The formula for the binomial coefficient is:
- \( _nC_r = \frac{n!}{r!(n-r)!} \)
- 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.
Using the above formula, we get:
- \( _{25}C_0 = \frac{25!}{0!(25-0)!} = \frac{25!}{1 \times 25!} = 1 \)
Permutation and Combination
Permutations and combinations, though related, describe different ways to arrange or select objects.
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.
- 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.
- If order doesn't matter, use combinations.
- If order does matter, 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.