If in the expansion of (1 + x)^m(1 - x)^n,the | vedclass

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(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.Multiplying the

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$12$

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(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.Multiplying the terms,we get $1 + (m - n)x + [\frac{n(n - 1)}{2} - mn + \frac{m(m - 1)}{2}]x^2 + \dots$.Given the coefficient of $x$ is $3$,we have $m - n = 3$,so $n = m - 3$.Given the coefficient of $x^2$ is $-6$,we have $\frac{n^2 - n}{2} - mn + \frac{m^2 - m}{2} = -6$.Substituting $n = m - 3$ into the equation:$\frac{(m - 3)(m - 4)}{2} - m(m - 3) + \frac{m^2 - m}{2} = -6$.Multiplying by $2$: $(m^2 - 7m + 12) - 2(m^2 - 3m) + (m^2 - m) = -12$.$m^2 - 7m + 12 - 2m^2 + 6m + m^2 - m = -12$.$-2m + 12 = -12$.$-2m = -24$,which gives $m = 12$.

If in the expansion of $(1 + x)^m(1 - x)^n$,the coefficients of $x$ and $x^2$ are $3$ and $-6$ respectively,then $m$ is:

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