[b]So, the question is about possible number of teams, thus a combination problem--- nCr
In nCr, r is 4, so nC4..

What is n? - It is given in the choices, that is CHOICE + 4
so (9+4)C4, (8+4)C4, (7+4)C4,....(2+4)C4, (1+4)C4..

Also \(10<nC4<100\)...
1) \(nC4>10\)
Take n as the least (1+4) so 5C4=5, which is less than 10, so take next higher
take n as the 6,so \(6C4 = \frac{6*5}{2}=15\), greater than 10.. so possible
2)\(nC4<100\)
take n as 13, \(13C4 = \frac{13*12*11*10}{4*3*2}=715\), way more than 100 so take the middle value in the choice
n =9, so \(9C4 = \frac{9*8*7*6}{4*3*2}=126\), so >100
n=8, so \(8C4 = \frac{8*7*6*5}{4*3*2}=70\)

so n is from 6 to 8
if 4 are sent, the number of astronaut left can be (6-4), (7-4) or (8-4), thus 2, 3, or 4

Bump

This one took me >2min to complete just the math portions and compare them, even starting with letter C!

Can we expect this amount of math in an actual GRE question?

Is there a shortcut or this is this a bloody, raw combinatorics problem that requires computation?

No, in my opinion

It is just theoretically you can get it

Like RC:ETS says you can have a passage of up to 6 questions.

No one saw a passage, ever, with more of 4 questions at the very most

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

I don't think this is representative of the GRE. But we can reduce the amount of calculation by being alert to the possible answers.

For example, the calculation for 6! was done mentally and then I calculated on paper only for 8! and 9! was just easily calculated using previous calculation and result done for 8!

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