Proceeding heuristically,
It cannot be \(4\) chosen out of \(5\) astronauts, as it will give us only a pool of \(5\) astronauts which is less than \(10\)
With \(6\) astronauts you will get \(6\times5\) in the numerator and \(6-4=2!=2\) in the denominator, which is \(15\).
So we now got the lower limit of the pool size.
Now I will skip a pool of \(7\) astronauts and go for a pool of \(8\) astronauts to check if this is the upper limit
\(4\) out of \(8\) will give me \(\frac{8\times7\times6\times5}{4\times3\times2} = 70\) which is less than \(100\). {calculation as it looks after simplifying}
Now clearly when \(70\) is multiplied by \(9\) and divided by \(5\) (if you can mentally visualize the formula), I am sure it is going to be close to double, but definitely greater than \(100\). In fact the actual value is \(126\).
\(\frac{9\times8\times7\times6\times5}{5\times4\times3\times2}= 126\) ( you have just multiplied the previous fraction by \(9\) and divided it by \(5\))
So we discard it.
So the pools are \(6,7\) and \(8\).
Out of these we have taken out \(4\) astronauts.
So those remaining are \(2,3\) and \(4\).
The answer choices are
F, G and H
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