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Chapter 5: Problem 4

Find the value of \(\frac{9 !}{4 !}\).

Short Answer

Expert verified
Answer: The value of \(\frac{9 !}{4 !}\) is 15120.

Step by step solution

01

Understanding Factorial

Factorial is a mathematical term represented by the symbol "!", which means to multiply a given number by all the positive whole numbers smaller than the given number. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\).
02

Evaluating the Factorials

Now that we know the factorial concept, we can evaluate the factorials for both numbers in the expression. \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\) \(4! = 4 \times 3 \times 2 \times 1 = 24\)
03

Dividing the Factorials

Finally, we'll divide the two factorial values we calculated to find the answer. \(\frac{9 !}{4 !} = \frac{362880}{24} = 15120\) So, the value of \(\frac{9 !}{4 !}\) is 15120.

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Most popular questions from this chapter

Factorise (a) \(x^{2}+8 x-9\), (b) \(x^{2}+9 x-22\) (c) \(x^{2}+10 x+9\), (d) \(x^{2}+7 x+12\) (e) \(x^{2}-7 x+12\)

Remove the square brackets from $$ \frac{1}{\omega}\left[\frac{s \omega}{s^{2}+\omega^{2}}-s\right] $$

Factorise \(144 \alpha^{2}-49 \beta^{2}\)

Remove the brackets from the given expression: \(\frac{3}{4}\left(\frac{1}{2} x+7\right)\)

Make \(n\) the subject of the formula $$ J=\frac{n E}{n L+m} $$
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