selfdoubt-ME-testseries [closed]

Question

we define a new measure ,called GoldIndex(G,C).it takes two arguments as input namely a graph G and set of colors C respectively . the subroutine outputs an integer denoting the number of ways assigning colors to vertices in G such that at least two vertices in G have the same color.Let $k_n$ denote the complete graph having n vertices respectively  and C={red,green,blue ,yellow}.then the GoldIndex ($k_3,C$) will be equal to_

my attempt –

the number of ways assigning colors to vertices in G such that at least two vertices in G have the same color= two vertices have same colors + three vertices have same colors (because $K_3$)

two vertices have same colors=$\binom{4}{2}$*3  // first choosing two  colors out of four and then assigning these two colors on three vertices 

three vertices have same colors (because $K_3$)=$\binom{4}{3}$*1 //  first choosing three  colors out of four and then assigning these three  colors on three vertices so only one way

so total no. of ways =18+4 =22

i don’t know where m i going wrong ,please help me-

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i know their solution is correct but i want to verify my approach-

Comments

two vertices have same colors= $\overset{\text{choose one color}}{\binom{4}{1}}*\overset{\text{select any two vertices}}{\binom{3}{2}}*\overset{\text{left vertex can be colored with any 3 colors}}{3} = 36$

three vertices have same colors=$\overset{\text{choose one color}}{\binom{4}{1}}*\overset{\text{select all three vertices}}{\binom{3}{3}}=4$

total ways = 36+4=40

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thank you @ Shubhgupta i got my mistake.