Question

Find the coefficient of ${{a}^{3}}{{b}^{3}}c$ in the expansion of ${{(2a+b+3c)}^{7}}$.

Answer 1

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 ${{a}^{3}}{{b}^{3}}c$.

Complete step-by-step answer:
We first find the number of combinations of arranging 3 a, 3 b and 1c. We have 7 terms. Thus, in general if we have 7 different terms, the total combinations would be 7! =5040. However, if we have some object of the same type, the number of combinations would be less since now we have identical objects and they would be treated as the same objects. To explain, if we have n objects with a objects of same type and b objects of same type (for aFor the number of combinations for arranging 3 a, 3 b and 1c, we have,
= $\dfrac{7!}{3!3!1!}$ -- (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
=3360
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 $\dfrac{n!}{x!y!z!}\times ({{p}^{x}}{{q}^{y}}{{r}^{z}})$.