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My approach : $$T(r) = \binom{n}{r-1}(3x)^{n-r+1} (-2)^{r-1}$$ here $T(r)$ is the $r$th term of the binomial expansion of $(3x-2)^{21}$. $$T(r) = \binom{n}{r-1}(3)^{n-r+1} (-2)^{r-1}(x)^{n-r+1}$$ here $n = 21$ , so , $$T(r) = \binom{21}{r-1}(3)^{22-r} (-2)^{r-1}(x)^{n-r+1}$$ So we need $\binom{21}{r-1}(3)^{22-r}(-2)^{r-1}$ to be maximum. I am not able to go further, please help me in this method or provide another method to solve the question in title.

sbares
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Sameer Nilkhan
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    You have been a member for 3 months and asked 22 questions. I think it is time you start learning how to format math properly. It's a lot easier for us to help you if we can actually read what your question says. – Arthur Jan 05 '20 at 18:21
  • https://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-tfrac-n2-rcei – lab bhattacharjee Jan 05 '20 at 18:26

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Compare the term $3^a2^{21-a}\begin{pmatrix}21\\a\\\end{pmatrix}$ with $3^{a+1}2^{20-a}\begin{pmatrix}21\\a+1\\\end{pmatrix}$. The ratio is $$\frac{3(21-a)}{2(a+1)}.$$ Since this is greater than $1$ for $a\le 12$ the maximum value occurs when $a=12$ i.e. $$3^{12}2^{9}\begin{pmatrix}21\\12\\\end{pmatrix}.$$