Let say we just have 6 English letters, A, B, C, D, E, F and k = 3(Assuming I have red, green, black color) with 3c2 = we get just 3 combinations, i.e p1(red, green), p2(black, green) and p3(red, black) with these three pairs we can assign to 3 pair of letters p1(AA) , p2(BB), p3(CC) but wait what about other 3 more pairs CC, EE, FF they can be assigned 3 colored as follows(Red Red ) , (Green Green) and (Black Black), hence we can use same approach.
For our problem we have 26pairs and if k = 6 we can just have 6c2 = 15 can be colored and 6 more with same color for pair so this is invalid we could not color all 26 pairs.
with k = 7, we can color 7C2 = 21 and plus 7 more so we have covered all 26 pairs.
Therefore C is the answer