If $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$,then $x+y+z=$

The number of four-digit numbers formed by using the digits $0, 2, 4, 5$ (without repetition) which are not divisible by $5$ is:

Let $0 \leq r \leq n$. If ${ }^{n+1} C_{r+1} : { }^{n} C_{r} : { }^{n-1} C_{r-1} = 55 : 35 : 21$,then $2n + 5r$ is equal to:

If four letters are chosen from the letters of the word $ASSIGNMENT$ and are arranged in all possible ways to form $4$-letter words (with or without meaning),then the total number of such words that can be formed is:

In a relay race there are five teams $A, B, C, D$ and $E$. What is the probability that $A, B$ and $C$ are the first three to finish (in any order)? (Assume that all finishing orders are equally likely.)

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