The coefficient of $a^{8} b^{6} c^{4}$ in the expansion $(a + b + c)^{18}$ is

^{18} C_{14} \cdot^{14} C_{8}

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^{18} C_{10} \cdot^{10} C_{6}

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^{18} C_{6} \cdot^{12} C_{8}

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^{18} C_{4} \cdot^{14} C_{6}

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Solution

The correct options are
A $^{18} C_{14} \cdot^{14} C_{8}$
B $^{18} C_{10} \cdot^{10} C_{6}$
C $^{18} C_{6} \cdot^{12} C_{8}$
D $^{18} C_{4} \cdot^{14} C_{6}$
$(a + (b + c))^{18}$
$=^{18} C_{r} a^{18 - r} (b + c)^{r}$
$=^{18} C_{r}^{r} C_{k} a^{18 - r} b^{r - k} c^{k}$
Now
$18 - r = 8$
$r = 10$
and $r - k = 6$
$k = 10 - 6 = 4$
Hence the coefficient, becomes
$^{18} C_{10}^{10} C_{4}$
$=^{18} C_{10}^{10} C_{6}$
$= \frac{18!}{(4!) (6!) (8!)}$
Hence answer is all of the above.

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