Why isn't the solution 6X6x6x6?

Explanation:

Case I: All numbers are different i.e. using 1, 2, 3, 4, 5, and 6
Number of ways of selecting 4 numbers out of 6 = $^6C_4 = 15$
Number of ways in which we can arrange these = $4! = 24$
Total number of ways = $(15)(24) = 360$

Case II: two 5's and two different numbers i.e. using 1, 2, 3, 4, 6, 5, and 5
Number of ways of selecting two 5's numbers out of 5 = $^5C_2 = 10$
Number of ways in which we can arrange these = $\frac{4!}{2!} = 12$
Total number of ways = $(10)(12) = 120$

Case III: two 6's and two different numbers i.e. using 1, 2, 3, 4, 5, 6 and 6
Number of ways of selecting two 6's numbers out of 5 = $^5C_2 = 10$
Number of ways in which we can arrange these = $\frac{4!}{2!} = 12$

Great explanation, thank you. Just quick note, the last case with repeating 5s and 6s is 4!/2!2! which is 12, so the total for A is 612
Total number of ways = $(10)(12) = 120$

Case IV: two 5's and two 6's i.e. 5, 5, 6 and 6
Number of ways in which we can arrange these = $\frac{4!}{(2!)(2!)} = 6$

Col. A: 360 + 120 + 120 + 6 = 606
Col. B: 560

Hence, option A