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Algebra II : Algebra II

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Example Questions

Example Question #51 : Multiplying And Dividing Factorials

$\frac{3!}{5!}\cdot \frac{6!}{7!}\div \frac{8!}{6!}$

Possible Answers:

$\frac{1}{7840}$

$\frac{19}{5}$

$\frac{5}{19}$

$\frac{1}{5600}$

Correct answer:

$\frac{1}{7840}$

Explanation:

Simplify the factorials in each fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form.

$\frac{3!}{5!}\cdot \frac{6!}{7!}\div \frac{8!}{6!}$

$\frac{3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1}\cdot \frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\div \frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$\frac{1}{5\cdot 4}\cdot \frac{1}{7}\div \frac{8\cdot 7}{1}$

$\frac{1}{20}\cdot \frac{1}{7}\div \frac{56}{1}$

$\frac{1}{140}\div \frac{56}{1}$

$\frac{1}{140}\cdot \frac{1}{56}$

$\frac{1}{7840}$

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Example Question #55 : Factorials

$\frac{20!\cdot 10!}{25!\cdot 5!}$

Possible Answers:

13080

$\frac{9}{13056}$

4080

$\frac{6}{1265}$

$\frac{1}{280795}$

Correct answer:

$\frac{6}{1265}$

Explanation:

Simplify the factorials the fraction by canceling common factors in the numerator and denominator. Normally, it would help to write out the factorials in expanded form, but since these are larger factorials that is not feasible. The best approach to cancelling these is thinking that 20! will be completely gone by taking it out of 25! and 5! will be completely gone by taking it out of 10!

$\frac{20!\cdot 10!}{25!\cdot 5!}$

$\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{25\cdot 24\cdot 23\cdot 22\cdot 21}$

$\frac{30240}{6375600}$

Then reduce this to lowest terms.

$\frac{3024}{637560}=\frac{1512}{318780}=\frac{756}{159390}=\frac{378}{79695}=\frac{54}{11385}=\frac{6}{1265}$

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Example Question #56 : Factorials

$\frac{2!}{13!}\div \frac{3!}{15!}$

Possible Answers:

$\frac{1}{56}$

$\frac{3}{105}$

218

Correct answer:

Explanation:

Simplify the factorials in each fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form.

$\frac{2!}{13!}\div \frac{3!}{15!}$

$\frac{2\cdot 1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\div \frac{3\cdot 2\cdot 1}{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 }$

$\frac{1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}\div \frac{1}{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$

Since it is division of fractions change the division sign to multiplication and flip the second fraction

$\frac{1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}\cdot \frac{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{1}$

$\frac{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}$

$\frac{15\cdot 14}{3}$

$\frac{210}{3}$

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Example Question #57 : Factorials

$\frac{8!\cdot 7!\cdot 3!}{9!\cdot 7!}$

Possible Answers:

$\frac{2}{3}$

$\frac{1}{9}$

$\frac{1}{24}$

Correct answer:

$\frac{2}{3}$

Explanation:

Simplify the factorials in the fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form, but I would suggest cancelling out the 7! in the numerator and denominator first.

$\frac{8!\cdot 7!\cdot 3!}{9!\cdot 7!}$

$\frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1}{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$\frac{3\cdot 2\cdot 1}{9}$

$\frac{6}{9}$

Then reduce the fraction to lowest terms

$\frac{2}{3}$

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Example Question #58 : Factorials

$\frac{4!}{6!}\cdot \frac{7!}{2!}\cdot 2!$

Possible Answers:

112

168

$\frac{1}{1080}$

Correct answer:

168

Explanation:

Simplify the factorials in the fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form.

$\frac{4!}{6!}\cdot \frac{7!}{2!}\cdot 2!$

$\frac{4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\cdot \frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1}\cdot 2\cdot 1$

$\frac{1}{6\cdot 5}\cdot \frac{7\cdot 6\cdot 5\cdot 4\cdot 3}{1}\cdot 2\cdot 1$

Combine each of the fractions into one fraction.

$\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5}$

$\frac{7\cdot 4\cdot 3\cdot 2\cdot 1}{1}$

$\frac{168}{1}$

168

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Example Question #51 : Multiplying And Dividing Factorials

Divide the factorials: $\frac{3!}{5!} \div \frac{5!}{3!}$

Possible Answers:

$\frac{1}{225}$

$\frac{1}{400}$

$\textup{The answer is not given.}$

$\frac{1}{5}$

Correct answer:

$\frac{1}{400}$

Explanation:

Dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

$\frac{3!}{5!} \div \frac{5!}{3!}=\frac{3!}{5!}\times \frac{3!}{5!}$

Write out the terms of the factorials.

$\frac{3!}{5!}\times \frac{3!}{5!}= \frac{3\times 2\times 1}{5\times 4\times 3\times 2\times 1}\times \frac{3\times 2\times 1}{5\times 4\times 3\times 2\times 1}$

Cancel all common terms.

$\frac{1}{5\times 4}\times\frac{1}{5\times 4} = \frac{1}{20}\times \frac{1}{20} = \frac{1}{400}$

The answer is: $\frac{1}{400}$

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Example Question #60 : Factorials

Divide the factorials: $\frac{2(4+1)!}{3!(2+3)!}$

Possible Answers:

$\frac{1}{720}$

504

$\frac{1}{3}$

$\frac{504}{505}$

432

Correct answer:

$\frac{1}{3}$

Explanation:

Evaluate the terms in parentheses first. Do not distribute the integer through the parentheses or this will change the value of the factorial!

$\frac{2(4+1)!}{3!(2+3)!} =\frac{2(5)!}{3!(5)!}$

The common terms in the numerator and denominator can be simplified.

$\frac{2}{3!} = \frac{2}{3 \times 2\times 1} = \frac{1}{3}$

The answer is: $\frac{1}{3}$

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Example Question #61 : Factorials

Divide the factorials: $\frac{3(2+3!)}{1-4!}$

Possible Answers:

$-\frac{360}{23}$

$-\frac{24}{23}$

$-\frac{87}{23}$

$\textup{Cannot be solved.}$

Correct answer:

$-\frac{24}{23}$

Explanation:

Simplify the factorials first.

$3! = 3\cdot2\cdot1 = 6$

$4! = 4\cdot3\cdot2\cdot1 = 24$

Replace the values back into the problem.

$\frac{3(2+3!)}{1-4!} = \frac{3(2+6)}{1-24} = \frac{24}{-23}$

The answer is: $-\frac{24}{23}$

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Example Question #201 : Mathematical Relationships And Basic Graphs

Simplify: $\frac{2(3!+4)}{(2+2)!}$

Possible Answers:

$\frac{181}{6}$

$\frac{5}{6}$

$\frac{10}{3}$

$\frac{91}{3}$

Correct answer:

$\frac{5}{6}$

Explanation:

Simplify the denominator.

$\frac{2(3!+4)}{(2+2)!}=\frac{2(3!+4)}{4!}$

Rewrite each factorial in the given fraction. Do not distribute the two through the quantity or this will change the value of the factorial.

$\frac{2(3!+4)}{4!} =\frac{2([3\times 2\times 1]+4)}{[4\times3\times 2\times 1] } = \frac{2(6+4)}{24}=\frac{2(10)}{24}=\frac{20}{24}$

Reduce this fraction.

The answer is: $\frac{5}{6}$

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Example Question #61 : Multiplying And Dividing Factorials

Divide the factorials: $\frac{2(3+2)!}{3!}$

Possible Answers:

120

604800

Correct answer:

Explanation:

To simplify this, we will need to evaluate the parentheses first. Do not distribute the integer two into the binomial, or the factorial value will change.

$\frac{2(3+2)!}{3!}=\frac{2(5)!}{3!}$

Expand the factorials.

$\frac{2(5)!}{3!} = \frac{2(5\cdot 4\cdot 3\cdot 2\cdot 1)}{3\cdot 2\cdot 1}$

Simplify the top and bottom terms.

$2(5\cdot 4) = 40$

The answer is:

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #51 : Multiplying And Dividing Factorials

Example Question #55 : Factorials

Example Question #56 : Factorials

Example Question #57 : Factorials

Example Question #58 : Factorials

Example Question #51 : Multiplying And Dividing Factorials

Example Question #60 : Factorials

Example Question #61 : Factorials

Example Question #201 : Mathematical Relationships And Basic Graphs

Example Question #61 : Multiplying And Dividing Factorials

All Algebra II Resources

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