Question 144493
This problem might be easier to approach if you
ask the opposite question:
"How many ways can the hospital purchase 4 of the 10
sets and only get 1 of the defective sets or none of
the defective sets?"
If I don't get any defective sets, then there are
7 possible sets for the 1st one, 6 possible sets for
the 2nd one, 5 for the 3rd, and 4 for the 4th
The number of different combinations is {{{a = 7*6*5*4}}}
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If I know I'm going to have 1 defective set, that means
I don't have 4 to choose from for the last set, I only
have the 3 defective ones, so the number of combinations
with 1 defective set is {{{b = 7*6*5*3}}}
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Then I have to ask, "If I include ALL the possible combinations
whether a set is defective or not, how many combinations
are there? That answer is: {{{c = 10*9*8*7}}}
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Now if I add {{{a}}} and {{{b}}}, I get the number of combinations
that I DON'T want {{{a + b = 7*6*5*4 + 7*6*5*3}}}
{{{a + b = 840 + 630}}}
{{{a + b = 1470}}}
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My answer should be {{{c - (a + b)}}}
{{{c = 10*9*8*7}}}
{{{c = 5040}}}
{{{5040 - 1470 = 3570}}}
If I'm right, there are 3570 different combinations of 4 sets that
include at least 2 defective ones. As with any problem like this,
reasoning can off, so see if you agree. 
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I thought this one over, and decided that I solved
for all possible ARRANGEMENTS of the TVs in question, as if it
made a difference which came 1st, 2nd, etc, which it doesn't
So change {{{a}}} to
{{{(7*6*5*4)/(4*3*2*1) = 35}}}
Change {{{b}}} to
{{{((7*6*5)/(3*2*1))*3 = 105}}}
Change {{{c}}} to
{{{(10*9*8*7)/(4*3*2*1) = 210}}}
{{{210 - (105 + 35) = 70}}}
There are 70 different combinations of 4 sets that
include at least 2 defective ones.
Of course, unless I'm wrong again. Check the answer.