${ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4=$

The number of ways in which $4$ different things can be distributed to $6$ persons so that no person gets all the things is

All five-letter words are made using all the letters $A, B, C, D, E$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_{n}$. Let the probability $P(W_{n})$ of choosing the word $W_{n}$ satisfy $P(W_{n}) = 2P(W_{n-1}), n > 1$. If $P(CDBEA) = \frac{2^{\alpha}}{2^{\beta}-1}, \alpha, \beta \in N$,then $\alpha + \beta$ is equal to

The last two digits of $2015! + 3^{2015}$ are:

There are three sections in a question paper,each containing $4$ questions. If a candidate has to answer only $5$ questions from this paper without leaving any section,then the number of ways the candidate can make the choice of questions is:

The value of ${}^{47}C_4 + \sum_{j=1}^5 {}^{(52-j)}C_3$ is