Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

Note that 7^r (r  N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
Thus, 7^m + 7ⁿ cannot end in 5 for any values of m, n  N. In other words, for 7^m + 7ⁿ to be divisible by 5, it should end in 0.
For 7^m + 7ⁿ to end in 0, the forms of m and n should be as follows :
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just 25 values of n for which 7^m + 7ⁿ ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
 There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^m + 7ⁿ is divisible by 5.
Hence

(1) Total number of ways of arranging m things = m!.
To find the number of ways in which p particular things are together, we consider p particular things as a group.
 Number of ways in which p particular things are together = (m – p + 1)! p!
So, number of ways in which p particular things are not together
= m! – (m – p + 1)! p!
Total number of ways =
Hence, both of statements are correct.

Let ‘l’ is associated with ‘r’ ,
r  {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
= =
=
= = 35
Hence (a) is correct answer.

|x|  k  –k  x  k ….(1)
& |y|  k  –k  y  k ….(2)
& |x – y|  k  |y – x|  k ….(3)

 – k  y – x  k  x – k  y  x + k
Number of points having integral coordinates
= (2k + 1)² – 2[k + (k – 1) + …. + 2 + 1]
= (3k² + 3k + 1).

The required no. of ways = no. of solution of the equation (x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = K)
Where 0  x_i 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^K in (1 + x + x² +….. + x⁹)⁶
= Coefficient of x^K in
= Coefficient of x^k in (1 – 6x¹⁰ + 15 x²⁰ – ….)
(1 + 6 C₁x + 7 C₂ x² + …. +(7 – K – 10 – 1) C_K–10 x^K–10 + ….+(7 + K – 1) C_K x^K + …)
= ^{k + 6}C_K – 6. ^K–4C_K–10
= ^{k + 6}C₆ – 6. ^K–4C₆ .

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is ^{m + n + k}C₃. But out of these m + n + k points, m points lie on I₁, n points lie on I₂ and k points lie on I₃ and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
^{m + n + k}C₃ – ^mC₃ – ⁿC₃ – ^kC₃.

A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.
We have given: the tangents drawn from a point on the hyperbola  to the ellipse  make angles α and β with the transverse axis of the hyperbola. 

An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
Now we have given the LCM as: r²t⁴s²
Consider following cases:
Case 1: if p contains r² then q will have r^k, for the value k=0,1.
So 2 ways.
Case 2: if q contains r² then p will have r^k, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r²
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.

Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m²n²
So it is m²n².