A) \[^{15}{{C}_{4}}\]
B) \[^{15}{{C}_{3}}\]
C) \[^{15}{{C}_{2}}\]
D) \[^{15}{{C}_{5}}\]
Correct Answer: A
Solution :
\[{{T}_{r+1}}={}^{15}{{C}_{r}}{{({{x}^{4}})}^{15-r}}{{\left( \frac{-1}{{{x}^{3}}} \right)}^{r}}\] \ \[{{T}_{r+1}}={}^{15}{{C}_{r}}\frac{{{(x)}^{60-4r}}{{(-1)}^{r}}}{{{(x)}^{3r}}}\]\[={}^{15}{{C}_{r}}{{(-1)}^{r}}{{(x)}^{60-7r}}\] Now putting \[60-7r=32\] Þ \[60-32=7r\] Þ \[r=\frac{28}{7}=4\] \[\therefore \] Coefficient of\[{{r}^{32}}={}^{15}{{C}_{4}}{{(-1)}^{4}}={}^{15}{{C}_{4}}\].
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