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Permutations & Combinations: There Four Colour Shifting Balls Colours Initially Ball

JEE Maths question with a full step-by-step solution.

Question
There are four colour-shifting balls of colours C1C_1, C2C_2, C3C_3 and C4C_4 initially. Each ball can change into any other colour or can regain its own colour at every second. The number of ways theircolours can change, if the maximum possible number of arrangements of all four balls is to be possible after 11 second, is equal to
Solution
Answer: 24
Step 1: Four balls can be arranged in the largest number of distinguishable orders exactly when their four colours are all different; if two balls share a colour, interchanging them produces no new arrangement and the count drops. So after one second the four balls must still carry the four colours C1,C2,C3,C4C_1,C_2,C_3,C_4, one each. Step 2: Each ball is assigned one of the four colours with all four colours used, which is a bijection from the four balls to the four colours, i.e. a permutation:
4!=244! = 24
 Step 3: Let DkD_k be the number of derangements of kk objects: D0=1D_0 = 1, D1=0D_1 = 0, D2=1D_2 = 1, D3=2D_3 = 2, D4=9D_4 = 9. Splitting the permutations by the number of fixed colours,
D4⏟none keeps+(41)D3+(42)D2+(43)D1+(44)D0\underbrace{D_4}_{\text{none keeps}}+\binom41D_3+\binom42D_2+\binom43D_1+\binom44D_0
 on calculating
9+4⋅2+6⋅1+4⋅0+1=9+8+6+0+1=249+4\cdot2+6\cdot1+4\cdot0+1 = 9+8+6+0+1 = 24
 the same total, as it must be, since ∑k(nk)Dn−k=n!\sum_k\binom nkD_{n-k} = n!. Answer: 2424
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