Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 10.1
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Exercise 52 Page 516

Practice makes perfect
a

Since we want to determine the number of combinations we can answer on our exam, we must use the following formula.

_nC_r=_nP_r/r! ⇔ _nC_r=n!/(n-r)!r! The exam consists of 12 questions and we want to answer 10 of them. This means we have n= 12 and r= 10.

_nC_r=n!/(n-r)!r!
_(12)C_(10)=12!/( 12- 10)! 10!
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Evaluate right-hand side
_(12)C_(10) = 12!/2! 10!

Write as a product

_(12)C_(10) = 12* 11* 10!/2!* 10!
_(12)C_(10) = 12* 11* 10!/2!* 10!
_(12)C_(10) = 12* 11/2!

2!=2

_(12)C_(10) = 12* 11/2
_(12)C_(10) = 132/2
_(12)C_(10) = 66

There are 66 ways we can pick 10 questions.

Alternative Solution

Using a Graphing Calculator
To calculate this we can also use the combinations formula on our graphing calculator. We start by entering the number of questions on the test.

Next, push the MATH button, scroll to PRB, and choose the third option. We finish by entering the number of questions we want to answer.

b

This time, the first three questions are mandatory. This means we want to calculate the number of combinations we can choose of 7 questions from a set of 9. Similarly as in Part A, we get the following result.

When the first three exercises are mandatory, there are 36 different ways we can pick the remaining 7 questions.