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 -

x₁ + x₂ + x₃ + x₄ + x₅ + x₆  17
When 1  x_i  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+7–1C_7–1 – ⁶C₁ × ^7+5–1C_7–1
= ¹⁷C₆ – 6 × ¹¹C₆ = 9604.

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+2–1}C_2–1
= ^25–3kC₁
= 25 – 3k
Hence number of solutions
= 25 × 9 – = 225 – 108 = 117.

Any divisor of 2⁵ · 3⁷ · 5³ · 7² is of the type of 2^l 3^m 5ⁿ 7^p, 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)
=
=.

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.

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

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

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=$r\cos \theta \; andy=r\sin \theta ,\; we\; get:$

 

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=$r\cos \theta \; andy=r\sin \theta ,\; we\; get:$