Question 1173140
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            I will help you with part  (2),  ONLY.

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<pre>
Actually, according to the problem's context, you select 2 defective calculators from 5 defective,


and then add to these 2 -- two other good calculators, selecting them from 12-5 = 7 good calculators.


The first  selection can be made in  {{{C[5]^2}}} = {{{(5*4)/2}}} = 10 ways.


The second selection can be made in  {{{C[7]^2}}} = {{{(7*6)/2}}} = 21 ways.


Hence, the whole selection can be made in 10*21 = 210 ways.    <U>ANSWER</U>
</pre>

Solved.



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