If binom{a^2+a}{3} = binom{a^2+a}{9},then a = | vedclass

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(A) Given the equation $\binom{a^2+a}{3} = \binom{a^2+a}{9}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,if $\binom{n}{x} = \binom{n}{y}$,then

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$3$

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(A) Given the equation $\binom{a^2+a}{3} = \binom{a^2+a}{9}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,if $\binom{n}{x} = \binom{n}{y}$,then either $x = y$ or $x + y = n$.Here,$3 
eq 9$,so we must have $3 + 9 = a^2 + a$.$a^2 + a = 12$.$a^2 + a - 12 = 0$.$(a + 4)(a - 3) = 0$.Thus,$a = 3$ or $a = -4$.Since $n$ in $\binom{n}{r}$ must be a non-negative integer and $n \geq r$,for $a = -4$,$n = (-4)^2 + (-4) = 16 - 4 = 12$,which is $\geq 9$. For $a = 3$,$n = 3^2 + 3 = 12$,which is $\geq 9$.Both values are mathematically valid,but based on the provided options,$a = 3$ is the correct choice.

If $\binom{a^2+a}{3} = \binom{a^2+a}{9}$,then $a = \dots$

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