A) 1
B) \[^{10}{{C}_{4}}\]
C) 4n
D) \[^{253}{{C}_{4}}\]
Correct Answer: B
Solution :
Given expansion \[{{(1+{{x}^{n}}+{{x}^{253}})}^{10}}\] Let \[{{x}^{1012}}={{(1)}^{a}}({{x}^{n}})b.{{({{x}^{253}})}^{c}}\] Here a, b, c, n are all +ve integers and \[a\le 10,b\le 10,c\le 4,n\le 22,\]\[a+b+c=10\] Now \[bn+253c=1012\] \[\Rightarrow \]\[bn=253(4-c)\] For \[c<4\]and \[n\le 22;b>10,\]which is not possible. \[\therefore \]\[c=4,b=0,a=6\] \[\therefore \]\[{{x}^{1012}}={{(1)}^{6}}.{{({{x}^{n}})}^{0}}.{{({{x}^{253}})}^{4}}\] Hence the coefficient of \[{{x}^{1012}}\] \[=\frac{10!}{6!0!4!}\]\[{{=}^{10}}{{C}_{4}}\]
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