Solution

Since, from the $$n^{th}$$ state, the electron may go to $$(n - 1)^{th}$$ state,$$(n - 2)^{th}$$ state ,... 2nd state or 1st state. So there are (n 1) possible transitions starting from the $$n^{th}$$ state. The atoms reaching $$(n - 1)^{th}$$ state may make (n 2) different transitions. The atoms reaching $$(n - 2)^{th}$$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy $$h\nu$$ and wavelength $$\lambda$$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted.
The total number of possible transitions are
$$(n-1), (n-2), (n-3), ................, 3, 2, 1 = \dfrac{n(n-1)}{2}$$

Therefore, for transition of an electron from higher energy state $$n = 4$$ to lower energy state $$n = 1$$ the number of photons emitted are
$$\dfrac{n(n-1)}{2} = \dfrac{4(4-1)}{2} = \dfrac{12}{2} = 6$$

Here, the electrons of each hydrogen atom are in same state hence in transition will emit photons of equal wavelengths for each transition hence, the maximum spectral lines emitted are $$6$$.