The number of ways we can select 3 cracked and 2 good is 1200 ways.
Given there are 5 eggs in which 3 eggs are cracked and 2 eggs which are good.
We have to find the number of ways we can select 3 cracked and 2 good eggs from 5 eggs.
Number of ways can be found through permutations.
The factorial of n equals the product of n with the smaller factorial..
Permutations is the technique of calculating various ways in which objects from a set may be selected, generally without replacement, to form subsets.
n[tex]P_{r}[/tex]=n!/(n-r)!
So the number of ways will be 5[tex]P_{3}[/tex]*5[tex]P_{2}[/tex]
([tex]5P_{3}[/tex] for cracked eggs and [tex]5C_{2}[/tex] for good eggs)
=5!/(5-3)!*51/(5-2)!
=5*4*3*2!/2!*5*4*3!/3!
=5*4*3*5*4
=1200
Hence there are 1200 ways through which 3 cracked and 2 good eggs can be selected from 5 eggs.
Learn more about permutations at https://brainly.com/question/1216161
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