List of top Mathematics Questions on permutations and combinations asked in KEAM

Let \( f(x) = ax^3 + bx^2 + cx + d \). If \( f \) has a local maximum value 21 at \( x = -1 \) and a local minimum value 7 at \( x = 1 \), then \( f(0) \) is equal to:
Let \( f(x) = \log_e \left( \frac{x^2 + 30}{11x} \right) \), for \( x \in [5,6] \). Then the point \( c \in (5,6) \) at which \( f'(c) = 0 \) is:
Let \[ g(x) = 4x + 3 \quad {and} \quad f(g(x)) = x^2 + 9. \] Then the value of \( f(7) \) is equal to
For any \(n≥0\) the value of \(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\) is 
Suppose there are 5 alike dogs, 6 alike monkeys and 7 alike horses. The number of ways of selecting one or more animals from these is
The number of ways in which we can distribute n identical balls in k boxes is 
If 11Pr=7920 and 11Cr = 330, then the value of r is equal to
The number of numbers greater than 6000 that can be formed from the digits 3, 5, 6, 7 and 9 (no digit is repeated in a number) is equal to
The number of arrangements containing all the seven letter of the word ALRIGHT that begins with LG is
The number of ways a committee of 3 women and 5 men can be formed from a panel of 8 men and 5 women is
If p and q are positive integers such that \(^{p+q}P_2=42\ and\ ^{p-q}P_2=20\), then the values of p and q respectively
The number of 3-digit numbers that can be formed from the digits 0, 2, 3, 5, 7 is (repetition is allowed)
How many numbers with no more than three digits can be formed using only the digits $1$ through $7$ with no digit used more than once in a given number?
There are $10$ persons including $3$ ladies. A committee of $4$ persons including at least one lady is to be formed. The number of ways of forming such a committee is
If $^{n}C_{r-1}=28$, $^{n}C_{r}=56$ and $^{n}C_{r+1}=70$, then the value of $r$ is equal to
The number of words that can be formed by using all the letters of the word $PROBLEM$ only one is
How many four digit numbers $abcd$ exist such that $a$ is odd, $b$ is divisible by $3$, $c$ is even and $d$ is prime?