For a Binomial distribution, B(n,p), where p+q=1, the sum and product of mean and variance are 8 and 12 respectively, when the value of n is:

For a Binomial distribution $B(n,p)$ where $p+q=1$:

Mean $= np$

Variance $= npq$

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Given that sum of mean and variance $= 8$ and product of mean and variance $= 12$.

Let mean $= np = m$

Then variance $= npq = mq$

The two conditions give:

$m + mq = 8$ ... (1)

$m \times mq = 12$ ... (2)

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From equation (1):

$m + mq = 8$

$m(1 + q) = 8$

From equation (2):

$m \times mq = 12$

$m^2q = 12$

$q = \dfrac{12}{m^2}$

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Substituting $q = \dfrac{12}{m^2}$ into $m(1 + q) = 8$:

$m\left(1 + \dfrac{12}{m^2}\right) = 8$

$m + \dfrac{12}{m} = 8$

$m^2 + 12 = 8m$

$m^2 - 8m + 12 = 0$

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Factoring the quadratic:

$(m - 6)(m - 2) = 0$

$m = 6$ or $m = 2$

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When $m = 6$:

$q = \dfrac{12}{36} = \dfrac{1}{3}$ (valid, since $0 < q < 1$)

$p = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

When $m = 2$:

$q = \dfrac{12}{4} = 3$ (invalid, since $q > 1$)

Therefore, $m = 6$ and $p = \dfrac{2}{3}$

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Since mean $= np = 6$ and $p = \dfrac{2}{3}$:

$n \times \dfrac{2}{3} = 6$

$n = 6 \times \dfrac{3}{2}$

$n = 9$

Therefore, the value of $n$ is $9$.