Let $R$ be a relation defined on the set $\{1,2,3,4\} \times\{1,2,3,4\}$ by

$$ \mathrm{R}=\{((a, b),(c, d)): 2 a+3 b=3 c+4 d\} . $$

Then the number of elements in R is

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^k}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to

Let $729,81,9,1, \ldots$ be a sequence and $\mathrm{P}_n$ denote the product of the first $n$ terms of this sequence.

If $2 \sum\limits_{n=1}^{40}\left(\mathrm{P}_n\right)^{\frac{1}{n}}=\frac{3^\alpha-1}{3^\beta}$ and $\operatorname{gcd}(\alpha, \beta)=1$, then

$\alpha+\beta$ is equal to