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# Find the coefficient of ${{a}^{3}}{{b}^{3}}c$ in the expansion of ${{(2a+b+3c)}^{7}}$.  Verified
Hint: For solving the question, we make use of concepts of binomial theorem along with permutations and combinations. Firstly, we try to find the number of ways in which we can arrange 3 a, 3 b and 1 c (since we have to find the coefficient of ${{a}^{3}}{{b}^{3}}c$). Next, we multiply this by the coefficient in the expression ${{(2a+b+3c)}^{7}}$ corresponding to ${{a}^{3}}{{b}^{3}}c$.
= $\dfrac{7!}{3!3!1!}$ -- (1)
This is for the term ${{a}^{3}}{{b}^{3}}c$. However, in the problem, we have ${{(2a+b+3c)}^{7}}$. Thus, we would have ${{(2a)}^{3}}{{b}^{3}}(3c)$ term. Thus, apart from the number of combination from (1), we would have to multiply by ${{2}^{3}}\times {{1}^{3}}\times 3$=24 [since, ${{(2a)}^{3}}{{b}^{3}}(3c)=({{2}^{3}}({{1}^{3}})(3))({{a}^{3}}{{b}^{3}}c)$, thus we also have to multiply by ${{2}^{3}}\times {{1}^{3}}\times 3$]. Now, finally, we combine this result with (1), we have,
= $\dfrac{7!}{3!3!1!}$$\times$24
Hence, the coefficient of ${{a}^{3}}{{b}^{3}}c$ is 3360.
Note: Finding the coefficient in a binomial theorem expansion requires a proper knowledge about permutations and combinations along with necessary expansion properties of the binomial theorem. In solving questions related to the above problem, we have a general formula to find the coefficient, we have, ${{(pa+qb+rc)}^{n}}$. To find, coefficient of ${{a}^{x}}{{b}^{y}}{{c}^{z}}$, the formula is $\dfrac{n!}{x!y!z!}\times ({{p}^{x}}{{q}^{y}}{{r}^{z}})$.