My Math Forum Probability(combination)

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 April 11th, 2011, 07:12 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability(combination) 1.How many diagonals does an octagon have? 2.16 students are divided into 3 groups,with group A having 9 students,group B having 4 students with the rest in group C.Calculate the number of ways to form a)the 3 groups b)the 3 groups with 4 particular students who must be in group A I don get it how question 1 relate to combination.please help me and question 2 too.thx
April 11th, 2011, 11:15 AM   #2
Math Team

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Re: Probability(combination)

Hello, hoyy1kolko!

Quote:
 1.How many diagonals does an octagon have?

$\text{An octagon has 8 vertices and 8 sides.}$

$\text{Select 2 of the vertices and connect them with a line segment.}$

$\text{There are: }\:_8C_2 \:=\:{8\choose2} \:=\:\frac{8!}{2!\,6!} \:=\:28\text{ ways.}$

$\text{All of them are diagonals }except\text{ the eight that form the sides of the octagon.}$

$\text{Therefore, there are: }\:28\,-\,8\:=\:20\text{ diagonals.}$

Quote:
 2. 16 students are divided into 3 groups, with group A having 9 students, group B having 4 students, with the rest in group C. Calculate the number of ways to form: [color=beige]. . [/color]a) the 3 groups.

$\text{This is a }partition.$ [color=beige] .[/color][color=red]**[/color]

$\text{The number of partitions is: }\:{15\choose 9,4,3} \:=\:\frac{16!}{9!\,4!\,3!} \:=\:400,400$

Quote:
 b) the 3 groups with 4 particular students who must be in group A.

$\text{Place the 4 particular students in group A.}$

$\text{Then we partition the remaining 12 students into groups of 5, 4 and 3.}$

$\text{There are: }\:{12\choose 5,4,3} \:=\:55,440\text{ ways.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[color=red]**[/color]

If you are not familiar with Partitions, you can still reason out the answer.

$\text{Select 9 of the 16 students to be in group A: }\:{16\choose9} \:=\:11,440\text{ ways.}$

$\text{Select 4 of the remaining 7 students to be in group B: }\:{7\choose4} \:=\:35\text{ ways.}$

$\text{The remaining 3 students will form group C: }\:1\text{ way.}$

$\text{Therefore, there are: }11,440\,\times\,35\,\times\,1 \:=\:400,400\text{ ways.}$

 April 12th, 2011, 02:33 AM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability(combination) but the answer provided for question 2(b) is 27720.

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