/*! 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 24 Evaluate each factorial expressi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate each factorial expression. $$\frac{18 !}{16 !}$$

Short Answer

Expert verified
The answer is \(306\).

Step by step solution

01

Understand Factorial Properties

Remember the property of factorials: n! = n*(n-1)!. You can use this property to simplify the factorial expression. In this expression, \(18!\) means the product of all positive integers from 18 to 1 and \(16!\) means the product of all positive integers from 16 to 1. So, \(16!\) is a part of \(18!\).
02

Simplify the Expression

Rewrite \(18!\) as \(18 \times 17 \times 16!\). Hence the given expression can be written as \(\frac{18 \times 17 \times 16!}{16!}\). You can see that \(16!\) appears in both the numerator and the denominator, so they cancel out. Hence, the given expression simplifies to \(18 \times 17\).
03

Evaluate the Result

Now, simply evaluate \(18 \times 17\) to find the final answer.

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Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

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