Counting absorption and emission lines for multi-level atoms

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YMMMA
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Homework Statement


If an atom in energy level 1, how many absorption spectral lines can an atom exhibit due to transition between 3 levels? How about 4 and 5 levels? And also in emission?

Homework Equations


Concept

The Attempt at a Solution


I am confused one book says it can absorb a photon to the third level at once, or to n=2, so there is only 2 possible transitions. Another book says it can absorb once to n=3, n=2, and from n=2 to n=3, so there are 3 possible transitions in total. How am I supposed to count the possible transitions? Here’s a question I encountered, but I need to know how to solve any problems of this kind. Any help is appreciated, Thanks in advance!
 

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YMMMA said:
one book says it can absorb a photon to the third level at once, or to n=2, so there is only 2 possible transitions. Another book says it can absorb once to n=3, n=2, and from n=2 to n=3, so there are 3 possible transitions in total.
It might come down to the exact wording of the statements. Can you post extracts?
 
I forgot to post the problem,too.
 

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Here they are.
 

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YMMMA said:
I forgot to post the problem,too.
The number of spectral lines are asked "if all transitions were possible ".
 
ehild said:
The number of spectral lines are asked "if all transitions were possible ".
So they are indeed three possible transitions.
 
YMMMA said:
So they are indeed three possible transitions.
Yes.
 
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YMMMA said:
Here they are.
I see no inconsistency. Some of these texts are considering only transitions from a given level, not all transitions between the levels in the set.
 
So why in question 5 didn’t we include transitions from n=3 to n=2 and n=2 to n=1, and n=3 to n=1...
 
YMMMA said:
So why in question 5 didn’t we include transitions from n=3 to n=2 and n=2 to n=1, and n=3 to n=1...
Because that question specifies transitions from level 4.
This is not the same as asking for all emission lines, since that would include 3->2, 3->1 and 2->1.
 
Got the difference, thanks!