Maths-
General
Easy
Question
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-
- 6C3 × 4C2
- 4P3 × 4P3
- 4C2 × 4P3
- None of these
Hint:
We will first start by using the method of selecting r objects out of n objects that is
for finding the ways in which we can select two chairs for women and three for men. Then we will permute the men and women among themselves.
The correct answer is: None of these
DETAILED SOLUTION
Now, we have been given 8 chairs which are numbered from 1 to 8. Also, it has been given that women choose the chairs from amongst the chairs marked 1 to 4, and then men select from remaining chairs.
In total there are 2 women and three men who wish to occupy one chair each.
Now, we know the number of ways of selecting r objects among n is
. So, we have the ways in which we can choose two chairs among four numbered 1 to 4 is
and we can arrange the women then in 2! ways. Also, we have the ways of selecting 3 chairs among the rest 6 chairs is
and in them we can permute the men in 3! ways.
So, in total we have number of possible arrangements as,
2!
3!
Now, we know that
= 
Therefore, we have,
Total ways =

It is important to note that we have used a fact that =
. This can be understood as we know that
=
and
=
. So, substituting this we have
=
.
Related Questions to study
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

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

physics-
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
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

physics-General
In the pulley arrangement 
But
is in downward direction and in the upward direction
, 
Acceleration of centre of mass


But
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
Maths-
If one root of the equation
is reciprocal of the one of the roots of equation
then
Let one of the roots of the equation
be
.
Then as per question one of the roots of the equation
will be
.


Multiplying equation 1 by
and equation 2 by a, then subtracting, we get:


putting the value in equation 1

Then as per question one of the roots of the equation
Multiplying equation 1 by
putting the value in equation 1
If one root of the equation
is reciprocal of the one of the roots of equation
then
Maths-General
Let one of the roots of the equation
be
.
Then as per question one of the roots of the equation
will be
.


Multiplying equation 1 by
and equation 2 by a, then subtracting, we get:


putting the value in equation 1

Then as per question one of the roots of the equation
Multiplying equation 1 by
putting the value in equation 1
Maths-
If the quadratic equation
and
have a common root then
is equal to
The two quadratic equations are 
and 
Let
be the common root.
Now,

subtracting the equations, we get

now putting the value in equation1

Let
Now,
subtracting the equations, we get
now putting the value in equation1
If the quadratic equation
and
have a common root then
is equal to
Maths-General
The two quadratic equations are 
and 
Let
be the common root.
Now,

subtracting the equations, we get

now putting the value in equation1

Let
Now,
subtracting the equations, we get
now putting the value in equation1
physics-
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14
to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,

Hence,
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14
to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

physics-General
At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,

Hence,
Maths-
In a Δabc if b+c=3a then
has the value equal to –
In a Δabc if b+c=3a then
has the value equal to –
Maths-General
Maths-
In a
simplifies to
In a
simplifies to
Maths-General