Maths-
General
Easy
Question
The number of ways in which n prizes can be distributed among n students when each student is eligible to get any number of prizes is-
- nn
- n!
- nn - n
- None of these
The correct answer is: n!
Related Questions to study
Maths-
In how many ways can six different rings be wear in four fingers?
Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× 4 ways of wearing 6 rings.
So, we have
= 4096 ways to wear 6 different types of rings.
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× 4 ways of wearing 6 rings.
So, we have
= 4096 ways to wear 6 different types of rings.In how many ways can six different rings be wear in four fingers?
Maths-General
Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× 4 ways of wearing 6 rings.
So, we have = 4096 ways to wear 6 different types of rings.
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× 4 ways of wearing 6 rings.
So, we have
= 4096 ways to wear 6 different types of rings.Maths-
How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?
How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?
Maths-General
Maths-
The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-
The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-
Maths-General
Maths-
Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-
Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-
Maths-General
Maths-
Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-
Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-
Maths-General
Maths-
A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span
DETAILED SOLUTION:
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are
,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).
So, a possible number of arrangements will be
Now as we know
So total number of arrangements is
= 210×(8! × 8!)
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are
, Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).
So, a possible number of arrangements will be
Now as we know
So total number of arrangements is
= 210×(8! × 8!)
A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span
Maths-General
DETAILED SOLUTION:
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).
So, a possible number of arrangements will be
Now as we know
So total number of arrangements is
= 210×(8! × 8!)
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are
, Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).
So, a possible number of arrangements will be
Now as we know
So total number of arrangements is
= 210×(8! × 8!)
Maths-
If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-
If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-
Maths-General
physics-
A thin uniform annular disc (see figure) of mass 
has outer radius 
and inner radius 
. The work required to take a unit mass from point 
on its axis to infinity is
Potential at point
will be obtained by in integration as given below. Let 
be the mass of small rings as shown
A thin uniform annular disc (see figure) of mass has outer radius and inner radius . The work required to take a unit mass from point on its axis to infinity is
physics-General
Potential at point
will be obtained by in integration as given below. Let be the mass of small rings as shownphysics-
The two bodies of mass 
and 
respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is
In the pulley arrangement 
But
is in downward direction and in the upward direction 
, 
Acceleration of centre of mass
But
is in downward direction and in the upward direction , Acceleration of centre of massThe two bodies of mass and respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is
physics-General
In the pulley arrangement
But is in downward direction and in the upward direction ,
Acceleration of centre of mass
But
is in downward direction and in the upward direction , Acceleration of centre of mass
maths-
If 
to 
terms, 
terms, 
, then 
If to terms, terms, , then
maths-General
Maths-
The coefficient of 
in 
is
The coefficient of in is
Maths-General
maths-
maths-General
chemistry-
Compounds (A) and (B) are – 
Compounds (A) and (B) are –
chemistry-General
Maths-
Maths-General
maths-
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:
maths-General
