Maths-
General
Easy
Question
Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -
- 109
- 118
- 119
- None of these
The correct answer is: 119
Required number of ways
D(r)

=
= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
= 10 + 20 + 45 + 44
= 119.
Related Questions to study
maths-
In how many ways can we get a sum of at most 17 by throwing six distinct dice -
x1 + x2 + x3 + x4 + x5 + x6 17
When 1
xi
6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
When 1
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
In how many ways can we get a sum of at most 17 by throwing six distinct dice -
maths-General
x1 + x2 + x3 + x4 + x5 + x6 17
When 1
xi
6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
When 1
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
maths-
The number of non negative integral solutions of equation 3x + y + z = 24
3x + y + z = 24, x
0, y
0, z
0
Let x = k
y + z = 24 – 3k …(1)
24 – 3k
0
k
8
0
k
8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 –
= 225 – 108 = 117.
Let x = k
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
The number of non negative integral solutions of equation 3x + y + z = 24
maths-General
3x + y + z = 24, x
0, y
0, z
0
Let x = k
y + z = 24 – 3k …(1)
24 – 3k
0
k
8
0
k
8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 –
= 225 – 108 = 117.
Let x = k
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
maths-
Sum of divisors of 25 ·37 ·53 · 72 is –
Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0
l
5, 0
m
7, 0 n 3 and 0
p
2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
.
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
Sum of divisors of 25 ·37 ·53 · 72 is –
maths-General
Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0
l
5, 0
m
7, 0 n 3 and 0
p
2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
.
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
maths-
The length of the perpendicular from the pole to the straight line
is
The length of the perpendicular from the pole to the straight line
is
maths-General
maths-
The condition for the lines
and
to be perpendicular is
The condition for the lines
and
to be perpendicular is
maths-General
Maths-
If f : R →R; f(x) = sin x + x, then the value of
(f-1 (x)) dx, is equal to
If f : R →R; f(x) = sin x + x, then the value of
(f-1 (x)) dx, is equal to
Maths-General
maths-
The polar equation of the straight line with intercepts 'a' and 'b' on the rays
and
respectively is
The polar equation of the straight line with intercepts 'a' and 'b' on the rays
and
respectively is
maths-General
maths-
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
maths-General
maths-
The polar equation of
axy is
The polar equation of
axy is
maths-General
maths-
If x, y, z are integers and x
0, y
1, z
2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -
Let y = p + 1 and z = q + 2.
Then x
0, p
0, q
0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 =
=
= 91.
Then x
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 =
If x, y, z are integers and x
0, y
1, z
2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -
maths-General
Let y = p + 1 and z = q + 2.
Then x
0, p
0, q
0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 =
=
= 91.
Then x
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 =
Maths-
If
then the equation whose roots are 
If
then the equation whose roots are 
Maths-General
Maths-
Let p, q
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
{1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q2 ≥ 4
q = 2,3,4
For p=2,q2≥8
q = 3,4
For p=3,q2≥12
q = 4
For p=4,q2≥16
q = 4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
Let p, q
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
{1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q2 ≥ 4
q = 2,3,4
For p=2,q2≥8
q = 3,4
For p=3,q2≥12
q = 4
For p=4,q2≥16
q = 4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
Maths-
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
Maths-
The cartesian equation of
is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
The cartesian equation of
is
Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
Maths-
The castesian equation of
is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
The castesian equation of
is
Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get: