/*! 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 50 (a) Evaluate \(\left(\begin{arra... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 50

(a) Evaluate \(\left(\begin{array}{c}11 \\ 4\end{array}\right)\).

Short Answer

Expert verified
To evaluate \(\left(\begin{array}{c}11\\\ 4\end{array}\right)\), we use the formula nCr = n! / (r!(n-r)!). After calculating the factorials and plugging in the values, we get: \(\left(\begin{array}{c}11\\\ 4\end{array}\right) = \frac{11!}{4!(11-4)!} = \frac{39916800}{(24)(5040)} = 330\). The answer is 330.

Step by step solution

01

Find the factorial of n (n!)

Find the factorial of 11: \(11! = 11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 = 39916800\).
02

Find the factorial of r (r!)

Find the factorial of 4: \(4! = 4\times3\times2\times1 = 24\).
03

Find the factorial of (n-r)

Let's calculate (n-r): \(11 - 4 = 7\). Now, find the factorial of 7: \(7! = 7\times6\times5\times4\times3\times2\times1 = 5040\).
04

Use the nCr formula to find the answer

Now that we have the factorials of n, r, and (n-r), we can use the nCr formula to find the solution: nCr = n! / (r!(n-r)!) So, \(\left(\begin{array}{c}11\\\ 4\end{array}\right) = \frac{11!}{4!(11-4)!} = \frac{39916800}{(24)(5040)} = 330\). The answer is 330.

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Key Concepts

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

Factorial
Factorials are a fundamental concept in combinatorics and essential for understanding combinations and binomial coefficients. A factorial, denoted with an exclamation mark (!), is the product of all positive integers up to a given number. For instance, the factorial of 5, written as \(5!\), is equal to \(5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow rapidly with larger numbers, making them significant in computations involving permutations and combinations.
  • Factorial of zero: \(0! = 1\), by definition.
  • Factorial notation is useful in various mathematical fields.
  • Factorials are key in calculating probabilities, sequences, and series.

Understanding factorials is crucial for solving problems like determining the number of ways to arrange a set of items or finding possible subsets.
Binomial Coefficient
Binomial coefficients are used to calculate the number of ways to choose a subset of elements from a larger set, where the order of selection does not matter. A binomial coefficient is often represented using the symbol \(\binom{n}{r}\), which is read as "n choose r." This represents the number of ways to choose \(r\) elements from a set of \(n\) elements.

The binomial coefficient can be calculated using the formula:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]
  • This formula uses factorials, making the understanding of factorials important.
  • The binomial coefficient is symmetric: \(\binom{n}{r} = \binom{n}{n-r}\).
  • They are used extensively in polynomial expansions, particularly in the binomial theorem.

In the given exercise, calculating \(\binom{11}{4}\) relies on the use of the factorial equation to enumerate the distinct subsets of 4 elements from a set of 11.
Combinations
Combinations are an important concept in mathematics for selecting items from a larger pool without regard to the order of selection. In simplest terms, a combination refers to the selection of items from a group where the order does not matter.
  • Combinations differ from permutations, where order does matter.
  • They are useful in scenarios where you want to know how many ways you can select items, such as drawing cards from a deck or choosing team members.
  • The formula used to calculate combinations, \(\binom{n}{r}\), directly ties to the concept of binomial coefficients.

By understanding combinations, you can solve various practical and theoretical problems related to selection and probability, as shown in the provided problem where we calculated the number of ways to choose 4 elements from a pool of 11, arriving at 330 distinct possibilities.

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