JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 16)
If the coefficient of a7b8 in the expansion of (a + 2b + 4ab)10 is K.216, then K is equal to _____________.
Answer
315
Explanation
$${{10!} \over {\alpha !\beta !\gamma !}}{a^\alpha }{(2b)^\beta }.{(4ab)^\gamma }$$
$${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$$
$$\alpha + \beta + \gamma = 10$$ ..... (1)
$$\alpha + \gamma = 7$$ .... (2)
$$\beta + \gamma = 8$$ ..... (3)
$$(2) + (3) - (1) \Rightarrow \gamma = 5$$
$$\alpha = 2$$
$$\beta = 3$$
so coefficients = $${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$$
$$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$$
$$ = 315 \times {2^{16}} \Rightarrow k = 315$$
$${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$$
$$\alpha + \beta + \gamma = 10$$ ..... (1)
$$\alpha + \gamma = 7$$ .... (2)
$$\beta + \gamma = 8$$ ..... (3)
$$(2) + (3) - (1) \Rightarrow \gamma = 5$$
$$\alpha = 2$$
$$\beta = 3$$
so coefficients = $${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$$
$$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$$
$$ = 315 \times {2^{16}} \Rightarrow k = 315$$
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