About Dr Nick

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Teaching Since:	May 2017
Last Sign in:	339 Weeks Ago, 5 Days Ago
Questions Answered:	19234
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Education

MBA (IT), PHD
Kaplan University
Apr-2009 - Mar-2014

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Professor
University of Santo Tomas
Aug-2006 - Present

Category > Statistics Posted 10 Nov 2017 My Price 13.00

solve this by breaking down the question into 2 smaller questions

I broke the question down into 2 parts, then solved them and multiplied the answer to get 11,880 ways.

We can solve this by breaking down the question into 2 smaller questions. The first question is "how many ways can we choose 4 chairs out of 12 to have someone sit in" and the second question is "how many ways can 4 people sit in 4 chairs". Our final answer will be the product of the 2 smaller questions.

The first smaller question - "how many ways can we choose 4 chairs out of 12 to have someone sit in" - is a combinations question (how many ways can we pick 4 chairs from a group of 12) - that is written in a few different notations - I'll write it out as Combination 12 choose 4 and in notation $C_{12, 4}$

For each combination of 4 chairs selected out of the 12, there are different ways the people can sit in them - we can have persons A B C D, or A B D C, etc. There are, in fact $4!$

ways each selection of chairs can be sat in.

Our final answer, therefore is

$(4!) (C_{12, 4})$

Let's solve

$(4!) = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

$(C_{12, 4}) = \frac{12!}{(4!) (8!)} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{(24) (8!)} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{24}$

Now let's combine the 2:

$(24) (\frac{12 \cdot 11 \cdot 10 \cdot 9}{24}) = (12 \cdot 11 \cdot 10 \cdot 9) = 11, 880$

Answers

Dr Nick

(4)

Status NEW Posted 10 Nov 2017 10:11 AM My Price 13.00

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