-aux.14.12 boys and 10 girls randomly selected

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SUMMARY

The problem involves selecting 2 boys from a group of 12 and 2 girls from a group of 10 to serve as crossing guards. The correct calculation uses combinations, specifically the binomial coefficient, resulting in 66 ways to choose the boys and 45 ways to choose the girls. Multiplying these outcomes yields a total of 2970 possible combinations. The initial incorrect approach using the fundamental counting principle led to an erroneous result of 14400.

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karush
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Of 12 boys and 10 girls, a teacher randomly selects 2 boys and 2 girls to be crossing guards, how many outcomes are there to this process?
Tried to the fundamental counting principal on this but 😢
$10\cdot10\cdot12\cdot12=14400$

Bk says its 2970?
 
Last edited:
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Hello, karush!

Of 12 boys and 10 girls, a teacher randomly selects 2 boys and 2 girls
How many outcomes are there to this process?
Select 2 boys from 12 boys.
\qquad There are: \:_{12}C_2 \,=\,{12\choose2}\,=\,\frac{12!}{2!\,10!} \,=\,66 ways.

Select 2 girls from 10 girls.
\qquad There are: \:_{10}C_2 \,=\,{10\choose2} \,=\,\frac{10!}{2!\,8!} \,=\,45 ways.

Therefore, there are: \:66\cdot45 \:=\:2970 outcomes.

 
Think I got it using combinations

$\frac{10!}{(10-2)!2!}\cdot\frac{12!}{\left(12-2\right)!2!}=2970$
 

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