My Math Forum Arrangement problems

 Probability and Statistics Basic Probability and Statistics Math Forum

 May 6th, 2015, 06:23 AM #1 Newbie   Joined: May 2015 From: hong kong Posts: 8 Thanks: 0 Arrangement problems The prime minister A,B,C,D,E,F and G of 7 countries will address at a summit meeting. (a) Find the number of arrangements that can be made so that (i) A will speak before C (ii) A will speak before C and C before E (b) In how many of those ways in (a ii) will C speak immediately after A ? Thanks!!
 May 6th, 2015, 07:20 AM #2 Senior Member     Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 (a i) $\displaystyle \binom{7}{2}\cdot 5!$ (a ii) $\displaystyle \binom{7}{3}\cdot 4!$
May 6th, 2015, 09:41 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, new hand!

Quote:
 The prime ministers of 7 countries {A,B,C,D,E,F,G} will address at a summit meeting. (a) Find the number of arrangements that can be made so that $\;\;\;\;$(i) A will speak before C

There are $7! \,=\,5040$ possible arrangements.

In half of them, A will precede C.
$\;\;\;\;\frac{1}{2}(5040) \,=\,2520$

Quote:
 (ii) A will speak before C and C before E.

I'll have to work on it . . .

Quote:
 (b) In how many of those ways in (a ii) will C speak immediately after A?

$\text{Duct-tape }A\text{ and }C\text{ together.}$
$\text{Then we have 6 "people" to arrange: }\:\left\{\boxed{AC},\,B, \,D,\,E,\,F,\,G\right\}$
There are: $\,6!= 720$ arrangements.

In half of them, $\boxed{AC}$ will precede $E.$
$\;\;\;\frac{1}{2}(720) \,=\,360.$

May 8th, 2015, 12:29 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, new hand!

Quote:
 The prime ministers of 7 countries {A,B,C,D,E,F,G} will address at a summit meeting. (a) Find the number of arrangements that can be made so that $\;\;\;\;$(ii) A will speak before C and C before E.

$\text{There are: }\,7!\,=\,5040\text{ possible orderings of the speakers.}$

$\text{In this list, }\{A,\,C,\,E\}\text{ appear in all }3!\,=\,6\text{ orderings.}$

$\text{W\!e want just }one\text{ order: }\,ACE.$

$\text{Therefore: }\,\frac{5040}{6} \:=\:840$

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