I think that 2 people from two people group can descend off from lift either on diff floor or same floor and same goes for the 3 and 4 groups also.
So, the groups are irrelevant, since people in each group can leave separately? That doesn't make much sense.
["they" refers to] 9 persons
I think the first "they" probably refers to the individuals (taking them as distinguishable), but the second "they" has to refer to the three groups, not to individuals (so that each
group leaves, together, on a different floor from the other
groups).
The next question for you is, since you know what you did is wrong, have you considered
changing your interpretation?
That is absolutely the thing you should have done, because quite often the most common mistakes in these problems are misinterpretation (or rather, choosing a different interpretation of a somewhat ambiguous problem than the author chose), and practical misinterpretation (using a counting scheme that doesn't really match the problem).
Here is how I would rephrase the question in the way that makes most sense to me, which is what I asked you to do:
In the Suniti Building in Mumbai there are 12 floors plus the ground floor. 9 people get into the lift of the building on the ground floor. The lift does not stop on the first floor. (That is, it can stop at any of floors 2 through 12.) If three groups of 2, 3 and 4 people alight from the lift on its upward journey, then in how many ways can they (the people) do so? (Assume the groups alight on different floors)
So I would interpret it as saying that the three groups have to leave on three different floors, so that two people leave on one floor, the three leave on another, and the four leave on yet another. So we can choose three floors, distinguishably (11P3), then choose two people for the group of 2 (9C2), and three for the group of three (7C3), and the remaining 4 are in the group of four (4C4 = 1). So the answer should be 11P3*9C2*7C3. Since I see this is not a choice, but 11P3*9C4*5C3 is, I see that they just chose the group of 4 first rather than last.
This is a risk in these multiple-choice problems that give a formula rather than a numerical answer, that you might use a different formula that is equivalent. Another way I considered was to treat the selection of people as arranging the letters AABBBCCCC, so that rather than combinations we would have 9!/(2!3!4!). But a glance at the choices had told me not to do that.
