Answer:
378
Step-by-step explanation:
ii. We use the binomial theorem which states that,
[tex](a + b) {}^{n} = \binom{n}{0} a {}^{n} b {}^{0} + \binom{n}{1} a {}^{n - 1} b {}^{1} + \binom{n}{2} a {}^{n - 2} b {}^{2} + \binom{n}{3} a {}^{n - 3} b {}^{3} + ...... \binom{n}{n - 2} a {}^{2} b {}^{n - 2} + \binom{n}{n - 1} {a}^{1} b {}^{n - 1} + \binom{n}{n} a {}^{0} {b}^{n} [/tex]
I'll explain in depth if you want me to.
We have
[tex](3a - \frac{b}{3} ) {}^{9} [/tex]
We must find the coeffeicent that include term.
[tex] {a}^{5} b {}^{4} [/tex]
The first term of the binomial expansion is a which has a degree 9 and b has a degree of 0. and since the term, a has a^5 and b^4. here this must that 9 must choose 4.
So let do the combinations formula,
[tex] \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(5 \times 4 \times 3 \times 2 \times 1)} [/tex]
[tex] \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} [/tex]
[tex] \binom{9}{4} = 126[/tex]
So during that point our expansion will include.
[tex]126(3a) {}^{5} ( - \frac{b}{3} ) {}^{4} [/tex]
[tex]126(243 {a}^{ {5} } \times \frac{b {}^{4} }{81} ) = 126(3a {}^{5} b {}^{4} )[/tex]
[tex]378 {a}^{5} {b}^{4} [/tex]
So the coeffeicent is 378