Algebra II : Multiplying and Dividing Factorials

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1 : Factorials

Stewie has \displaystyle \frac{5!}{3!}  marbles in a bag. How many marbles does Stewie have? 

Possible Answers:

\displaystyle 12

\displaystyle 2!

\displaystyle 6

\displaystyle 1.66

\displaystyle 20

Correct answer:

\displaystyle 20

Explanation:

\displaystyle 5!/3! = (5*4*3*2*1)/(3*2*1)

Simplifying this equation we notice that the 3's, 2's, and 1's cancel so

\displaystyle 5!/3! = 5*4 = 20

Alternative Solution

\displaystyle 5!/3! = (120/6) = 20

Example Question #2 : Factorials

Which of the following is NOT the same as \displaystyle \frac{6!\times5!}{4!}?

Possible Answers:

\displaystyle \frac{6\times(5!)^2}{4!}

\displaystyle 30\times5!

\displaystyle 5\times6!

\displaystyle 6\times5!

\displaystyle 6\times5^{2}\times4!

Correct answer:

\displaystyle 6\times5!

Explanation:

The \displaystyle 4! cancels out all of \displaystyle 6! except for the parts higher than 4, this leaves a 6 and a 5 left to multilpy \displaystyle 5!

Example Question #3 : Factorials

Simplify the following expression:

\displaystyle \frac{7!(n+3)!}{6!(n+2)!}

Possible Answers:

\displaystyle 7n +3

\displaystyle 7n + 21

\displaystyle 7!(n+3)

\displaystyle \frac{n+3}{n+2}

\displaystyle n+3

Correct answer:

\displaystyle 7n + 21

Explanation:

Recall that \displaystyle (n+3)! = (n+3)(n+2)(n+1)(n)....

Likewise, \displaystyle 6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1.

Thus, the expression \displaystyle \frac{7!(n+3)!}{6!(n+2)!} can be simplified in two parts:

\displaystyle \frac{7!}{6!} = \frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} = 7 

and

\displaystyle \frac{(n+3)!}{(n+2)!} = \frac{(n+3)(n+2)(n+1...)}{(n+2)(n+1)...} = n+3

The product of these two expressions is the final answer: \displaystyle 7n + 21

Example Question #1 : Factorials

\displaystyle \frac{6!}{2!4!} = ?

Possible Answers:

\displaystyle 1

\displaystyle 15

\displaystyle \frac{3}{4}

\displaystyle 12

\displaystyle 30

Correct answer:

\displaystyle 15

Explanation:

To simplify this, just write out each factorial:

\displaystyle \frac{6!}{2!4!} = \frac{6*5*4*3*2*1}{(2*1)(4*3*2*1)} = \frac{6*5}{2*1} = 15

Example Question #1 : Multiplying And Dividing Factorials

Find the value of:

\displaystyle \frac{7!}{3!}

Possible Answers:

\displaystyle 2520

\displaystyle 2.33

\displaystyle 210

\displaystyle 840

\displaystyle 4

Correct answer:

\displaystyle 840

Explanation:

The factorial sign (!) just tells us to multiply that number by every integer that leads up to it.  So, \displaystyle \frac{7!}{3!} can also be written as: 

\displaystyle \frac{(7\times6\times5\times4\times3\times2\times1)}{(3\times2\times1)}

To make this easier for ourselves, we can cancel out the numbers that appear on both the top and bottom:

\displaystyle 7\times6\times5\times4 = 840

Example Question #2 : Factorials

Which of the following is equivalent to \displaystyle 6!\times4!?

Possible Answers:

None of the other answers are correct.

\displaystyle 24!

\displaystyle 30\times(4!)^2

\displaystyle (6!)^{2}-4!

\displaystyle 10!

Correct answer:

\displaystyle 30\times(4!)^2

Explanation:

This is a factorial question. The formula for factorials is \displaystyle n! = n\times(n-1)\times(n-2)\times...\times1.

 \displaystyle 6!=6\times5\times4! = 30\times 4!

Example Question #5 : Multiplying And Dividing Factorials

Divide \displaystyle 2015! by \displaystyle 2014!

Possible Answers:

\displaystyle 3,870,890

\displaystyle 4,058,210

\displaystyle 2015

\displaystyle 1,000,000

\displaystyle 21,900,320

Correct answer:

\displaystyle 2015

Explanation:

A factorial is a number which is the product of itself and all integers before it. For example \displaystyle 5!=5\cdot 4 \cdot 3 \cdot 2 \cdot 1

In our case we are asked to divide \displaystyle 2015! by \displaystyle 2014!. To do this we will set up the following:

\displaystyle \frac{2015!}{2014!}

We know that \displaystyle 2015! can be rewritten as the product of itself and all integers before it or:

\displaystyle 2015!=2015 \cdot 2014!

Substituting this equivalency in and simplifying the term, we get:

\displaystyle \frac{2015 \cdot 2014!}{2014!}=2015

 

Example Question #1 : Factorials

\displaystyle \frac{\left ( n+2\right )!}{n!}

If \displaystyle n is a postive integer, which of the following answer choices is a possible value for the expression.

Possible Answers:

\displaystyle 30

\displaystyle 15

\displaystyle 24

\displaystyle 21

\displaystyle 104

Correct answer:

\displaystyle 30

Explanation:

This expression of factorials reduces to (n+1)(n+2).  Therefore, the solution must be a number that multiplies to 2 consecutive integers.  Only 30 is a product of 2 consecutive integers.  \displaystyle 5\ast6=30

So n would have to be 4 in this problem.

Example Question #1 : Multiplying And Dividing Factorials

Simplify:

\displaystyle \frac{20!}{17!\times 2!}

Possible Answers:

\displaystyle 1.18

\displaystyle 0.68

\displaystyle 1.05

\displaystyle 3420

\displaystyle 6840

Correct answer:

\displaystyle 3420

Explanation:

Remember what a factorial is, and first write out what the original equation means. A factorial is a number that you multiply by all whole numbers that come before it until you reach one.

You can simplify because all terms in the expression 17! are found in 20!.

Thus:

\displaystyle \frac{20!}{17!\times 2!}=\frac{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot 2}

\displaystyle =\frac{20\cdot19\cdot18}{2}=10\cdot19\cdot18=3420

Example Question #10 : Multiplying And Dividing Factorials

Simplify:  \displaystyle \frac{4!+1}{3!+1}

Possible Answers:

\displaystyle 4

\displaystyle \frac{5}{4}

\displaystyle 5

\displaystyle \frac{25}{7}

\displaystyle \frac{17}{10}

Correct answer:

\displaystyle \frac{25}{7}

Explanation:

Rewrite the factorials in multiplicative order.

\displaystyle \frac{4!+1}{3!+1} = \frac{(4\times 3\times2\times1)+1}{(3\times2\times1)+1}

In this scenario, the numbers of the factorial in the numerator and denominator CANNOT cancel.  Simplify by multiplying out the factorials.

\displaystyle \frac{(4\times 3\times2\times1)+1}{(3\times2\times1)+1}= \frac{24+1}{6+1}=\frac{25}{7}

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