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 May 16th, 2011, 04:28 PM #1 Senior Member   Joined: Mar 2011 Posts: 105 Thanks: 0 Permutation and Combination How many different way to arrange 12 stuffed animals on a counter ( 5 identical elephants and 7 identical monkeys )? How many different way to choose 5 from 12 monkey?? Why the answers of these question are same with one another !! I"M so confused
 May 16th, 2011, 04:58 PM #2 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Permutation and Combination Are we assuming that switching one monkey with another will still be the same combination?
May 16th, 2011, 05:08 PM   #3
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Re: Permutation and Combination

Quote:
 Originally Posted by The_Fool Are we assuming that switching one monkey with another will still be the same combination?
yes

May 16th, 2011, 05:12 PM   #4
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Re: Permutation and Combination

Quote:
 Originally Posted by daivinhtran How many different way to arrange 12 stuffed animals on a counter ( 5 identical elephants and 7 identical monkeys )?
If you count the possible arrangements of five elephants you also count the possible arrangements of seven monkeys, they fill in the "blank spaces", i.e. the spaces not occupied by the five elephants.

If a = b + c, a nCr b = a nCr c.

 May 16th, 2011, 05:19 PM #5 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Permutation and Combination In addition to to his explanation, there is a third way to find this. Take the total ways to arrange 12 stuffed animals, divide by the number of ways to arrange 7 monkeys, then divide again by the number of ways to arrange 5 elephants. This will eliminate combinations of monkeys and elephants that are switched with their own kind. This will also be the number of unique ways to arrange them.
May 16th, 2011, 08:59 PM   #6
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Re: Permutation and Combination

Hello, daivinhtran!

Quote:
 (a) How many different way to arrange 12 stuffed animals on a counter? [color=beige]. . [/color](5 identical elephants and 7 identical monkeys) (b) How many different way to choose 5 from 12 monkey?? Why are these answers the same?

(a) Consider that there are 12 spaces on the counter.

[color=beige]. . .[/color]Choose 5 spaces for the elephants.[color=beige] .[/color](The monkeys will go in the other 7 spaces.)

[color=beige]. . .[/color]$\text{There are: }\:_{12}C_5 \:=\:\frac{12!}{5!\,7!} \:=\:792\text{ choices for placing the elephants.}$

[color=beige]. . .[/color]Therefore, there are 792 ways to arrange the stuffed animals in a row.

(b) There are 12 monkeys and we will choose 5 of them.

[color=beige]. . .[/color]$\text{There are: }\:_{12}C_5 \:=\:\frac{12!}{5!\,7!} \:=\:792\text{ choices for the 5 monkeys.}$

In both cases, we are choosing 5 objects from a set of 12 available objects.

May 17th, 2011, 08:59 AM   #7
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Re: Permutation and Combination

Quote:
Originally Posted by soroban
Hello, daivinhtran!

Quote:
 (a) How many different way to arrange 12 stuffed animals on a counter? [color=beige]. . [/color](5 identical elephants and 7 identical monkeys) (b) How many different way to choose 5 from 12 monkey?? Why are these answers the same?

(a) Consider that there are 12 spaces on the counter.

[color=beige]. . .[/color]Choose 5 spaces for the elephants.[color=beige] .[/color](The monkeys will go in the other 7 spaces.)

[color=beige]. . .[/color]$\text{There are: }\:_{12}C_5 \:=\:\frac{12!}{5!\,7!} \:=\:792\text{ choices for placing the elephants.}$

[color=beige]. . .[/color]Therefore, there are 792 ways to arrange the stuffed animals in a row.

(b) There are 12 monkeys and we will choose 5 of them.

[color=beige]. . .[/color]$\text{There are: }\:_{12}C_5 \:=\:\frac{12!}{5!\,7!} \:=\:792\text{ choices for the 5 monkeys.}$

In both cases, we are choosing 5 objects from a set of 12 available objects.

I got it, now, thank you

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