$A$ question paper has $3$ parts $A, B, C$. Part $A$ contains $7$ questions,part $B$ contains $5$ questions,and part $C$ contains $3$ questions. If a candidate is allowed to answer not more than $4$ questions from part $A$,not more than $3$ questions from part $B$,and not more than $2$ questions from part $C$,then the number of ways in which a candidate can answer exactly $7$ questions is

Let $S$ denote the set of $4$-digit numbers $abcd$ such that $a > b > c > d$ and $P$ denote the set of $5$-digit numbers having the product of its digits equal to $20$. Then $n(S) + n(P)$ is equal to:

The remainder obtained when $1! + 2! + 95!$ is divided by $15$ is

In an examination,there are $5$ multiple choice questions with $3$ choices each,out of which exactly one is correct. There are $3$ marks for each correct answer,$-2$ marks for each wrong answer,and $0$ marks if the question is not attempted. The number of ways a student appearing in the examination can get exactly $5$ marks is:

Consider the following statements:
$(i)$ The number of one-one functions from set $A$ to set $B$,where $O(A) = m$ and $O(B) = n$ $(m \leq n)$,is given by ${}^n P_m$.
(ii) The number of ways in which $n$ people can be arranged at a circular table is $\frac{(n-1)!}{2}$.
(iii) The number of ways of selecting at least one thing out of the given $n$ distinct things is $2^n - 1$.
(iv) The number of ways in which $n$ distinguishable objects can be distributed into $k$ distinguishable bins is ${}^n C_{k-1}$.
Which of the following is true?

The sum of all the $4$-digit numbers formed by taking all the digits from ${0, 3, 6, 9}$ without repetition is