Question
If 9P5 + 5 9P4 =
, then r =
- 4
- 5
- 9
- 10
Hint:
Here we need to find the value of the given variable. Here we will first apply the formula of the permutation to simplify the equation. Then we will apply the mathematical operations that will be required to simplify the given expression further. After simplifying the terms, we will get the value of the required variable.
Formula used:

The correct answer is: 5
Detailed Solution
Here we need to find the value of the given variable.
The given expression is :


Related Questions to study
Assertion (A) :If
, then 
Reason (R) :
Assertion (A) :If
, then 
Reason (R) :
A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
If x is real, then maximum value of
is
If x is real, then maximum value of
is
The value of 'c' of Lagrange's mean value theorem for
is
The value of 'c' of Lagrange's mean value theorem for
is
The value of 'c' of Rolle's mean value theorem for
is
The value of 'c' of Rolle's mean value theorem for
is
The value of 'c' of Rolle's theorem for
–
on [–1, 1] is
The value of 'c' of Rolle's theorem for
–
on [–1, 1] is
For
in [5, 7]
For
in [5, 7]
The value of 'c' in Lagrange's mean value theorem for
in [0, 1] is
The value of 'c' in Lagrange's mean value theorem for
in [0, 1] is
The value of 'c' in Lagrange's mean value theorem for
in [0, 2] is
The value of 'c' in Lagrange's mean value theorem for
in [0, 2] is
The equation
represents
The equation
represents
The polar equation of the circle whose end points of the diameter are
and
is
The polar equation of the circle whose end points of the diameter are
and
is
The radius of the circle
is
The radius of the circle
is
The adjoining figure shows the graph of
Then –

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.
The adjoining figure shows the graph of
Then –

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.
Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.
Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.