- CFA Exams
- 2021 CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 10. Principles of Counting

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**CFA Practice Question**

A company has three different projects that must be assigned to members of its staff. There are 7 staff member being considered for these projects. How many ways are there to assign each project to one individual only?

B. 210

C. 840

A. 35

B. 210

C. 840

Correct Answer: B

Use the permutation formula to find how many ways one can combine 3 people out of 7. 7!/4! = 210

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**User Contributed Comments**
7

User |
Comment |
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batman |
Apparently this is a permutation problem, since the projects are different from one another. |

ylepape |
I agree. An empirically, there are 7 ways to make the first selection, 6 for the next and 5 for the last = 210 |

surob |
I guess the key word to pay attention to is "each project to one individual only", which tells us to use permutation. If it was "each individual to one project only", we would use combination... |

nsmwaura |
This is confusing me...I need some devine intervention soon.... i still can't get where to use Permutation or combination |

StanleyMo |
Hello nsmwaura, lets hope this make things simple: Q. How many different signals can be made by 5 flags from 8-flags of different colours? Ans. Number of ways taking 5 flags out of 8-flage = 8P5 = 8!/(8-5)! = 8 x 7 x 6 x 5 x 4 = 6720 We are using permutation as all the 8 colors matter instead of choosing 8 7 6 5 4. which means all the "parameter" are unique. and also, How many words can be made by using the letters of the word "SIMPLETON" taken all at a time? kind of this questions, hope this helps. |

viannie |
Permutation: when order does matter. In this problem, when the 1st project is assigned, there are 6 person left for the 2nd project assignment. Then by the 3rd project, there are 5 person left which only 1 will be selected to handle the project. So 7 x 6 x 5 = 7P3 Just remember that permutation cannot be smaller than combination. Combination can be any one irregardless of order but permutation is more fussy. |

kaichan91 |
Since the projects are stated to be different, would it make more sense to think of the problem by thinking 7P1 * 6P1 * 5P1 instead of 7P3? Or is it 7P3 because permutations already factor in the difference in projects? |