If ^{20}C_{n+2} = ^nC_{16},then the value of | vedclass

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(D) The given equation is $^{20}C_{n+2} = ^nC_{16}$.For a combination $^nC_r$ to be defined,the condition $n \ge r$ must be satisfied.In the expressio

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(D) The given equation is $^{20}C_{n+2} = ^nC_{16}$.For a combination $^nC_r$ to be defined,the condition $n \ge r$ must be satisfied.In the expression $^{20}C_{n+2}$,we must have $20 \ge n+2$,which implies $n \le 18$.In the expression $^nC_{16}$,we must have $n \ge 16$.Combining these,we get $16 \le n \le 18$.If $n = 16$,then $^{20}C_{18} = ^{16}C_{16} \implies 190 = 1$,which is false.If $n = 17$,then $^{20}C_{19} = ^{17}C_{16} \implies 20 = 17$,which is false.If $n = 18$,then $^{20}C_{20} = ^{18}C_{16} \implies 1 = 153$,which is false.Thus,there is no integer value of $n$ that satisfies the equation.

If $^{20}C_{n+2} = ^nC_{16}$,then the value of $n$ is

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