There are 5 places
― ― ― ― ―
Given: L   I L A C
The derangements formula ⎣n!/e⎦ cannot be directly performed as there are repeated characters.
Let’s proceed in regular manner:
The L, L can be placed in other ‘3’ places as

(1) Can be arranged such that A, I, C be placed in three positions excluding ‘C’ being placed at its own position, which we get only 2×2×1=4 ways.
Similarly (2) can be filled as A, I, C being placed such that 4th position is not filled by A, so we have 2×2×1= 4 ways. Similarly with (3).
Totally, we get 4+4+4 = 12 ways.

Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e., |A∪B∪C|
We know,
|A∪B∪C| = |A|+|B|+C-|A∩B|-|A∩C|-|B∩C|+|A∩B|
|A| = number of integers divisible by 3
[500/3 = 166.6 ≈ 166 = 166]
|B| = 100
[500/5 = 100]
|C| = 71
[500/7 = 71.42]
|A∩B| = number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋ ≈ 33
|A∩B| = 33
|A∩C| = 500/LCM(3,7) 500/21 = 23.8 ≈ 28
|B∩C| = 500/LCM(5,3) = 500/35 = 14.48 ≈ 14
|A∩B∩C| = 500/LCM(3,5,7) = 500/163 = 4.76 ≈ 4
|A∪B∪C| = |A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
= 166+100+71-33-28-14+4
= 271

a_n = number of n-bit strings, that do not have two consecutive 1’s.
If n=1, we have {0,1}
# Occurrences = 2
If n=2, we have {00,01,10}
# Occurrences = 3
If n=3, we have {000,001,010,100,101}
# Occurrences = 5
It is evident that a₃ = a₁ + a₂
Similarly, a_n = a_n-1 + a_n-2

Co-efficient of x¹² in (x³ + x⁴ + x⁵ + x⁶+...)³
⇒ [x³(1 + x + x² + x³ + ...)]³
= x⁹(1 + x + x² + x³ + ...)³
First Reduction:
As x⁹ is out of the series, we need to find the co-efficient of x³ in (1 + x + x² + ⋯)³

Here, m=3, k=3, the coefficient

= ⁵C₃ = 5!/2!3! = 10

a_n = 6n² + 2n + a_(n-1)
Replace a_(n-1)

⇒ a_n = 6n² + 2n + 6(n-1)² + 2(n-1) + 6(n-2)² + 2(n-2) + ⋯ a₁
Given that a₁ = 8, replace it
⇒ a_n = 6n² + 2n + 6(n-1)² + 2(n-1) + 6(n-2)² + 2(n-2) + ⋯8
= 6n² + 2n + 6(n-1)² + 2(n-1) + 6(n-2)² + 2(n-2) + ⋯ + 6(1)² + 2(1)

= 6(n² + (n-1)² + (n-2)² + ⋯ + 2² + 1²) + 2(n + (n-1) + ⋯1)
Sum of n² = (n(n+1)(2n+1))/6
Sum of n = (n(n+1))/2
= 6 × (n(n+1)(2n+1))/6 + 2×(n(n+1))/2
= n(n+1)[1+2n+1]
= n(n+1)[2n+2]
= 2n(n+1)²
Given a₉₉ = k×10⁴
a₉₉ = 2(99)(100)² = 198 × 10⁴
∴k = 198

Let N = 2100
=2²+3×5²×7 (i.e., product of primes)
Then the number of divisions of 2100 is
(2+1)∙(1+1)∙(2+1)∙(1+1) i.e., (3)(2)(3)(2) i.e., 36

Just try to write all the four digit numbers containing the digits 1, 2 and 3 only, in decreasing order.
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4-digit no. are possible.

No twos: 1111111111⇒ 1 pennant
Single two: 211111111 ⇒ 9!/8!1! = 9 pennants
Two twos: 22111111 ⇒ 8!/6!2! = 28
Three twos: 2221111 ⇒ 7!/3!4! = 35
Four twos: 222211 ⇒ 6!/4!2! = 15
Five twos: 22222 ⇒ 1
Total = 89 pennants.

First lets find prime factorization of 2014
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8

Lets name one unit up movement as 'u' and one unit right movement as 'r'.
So now we have 10 u's and 10 r's, i.e.,
uuuuuuuuuurrrrrrrrrr
So, finally the no. of arrangements of above sequences is,

So to solve this lets find the no. of paths possible if line segment from (4,4) to (5,4) is taken. And then we will subtract it from the total no. of paths possible.
So, no. of paths possible if line segment from (4,4) to (5,4) is taken is,
= paths possible from (0,0) to (4,4) * paths possible from (5,4) to (10,10)
= {uuuurrrr} * {uuuuuurrrrr}

Hence, the final answer is

Let c=1, x₀=1
Then,
x₁ = c ⋅ (x₀)² - 2 = 1 ⋅ (1)² - 2 = -1
x₂ = c ⋅ (x₁)² - 2 = 1 ⋅ (-1)² - 2 = -1
So, the value converges to -1, which is equal to

Exactly, only (B) is answer. As all the term of x converges to -1.

For the series to converge the limit: 'n' tends to infinity of (x_n+1 / x_n) should be <1.
From the recurrence we should have c(x_n)² - x_n - 2 < 0
For all the above values of c we have above equation as negative.

Given a set of elements N = {1, 2, 3, ...N}
Two arbitrary subsets A⊆N and B⊆N.
Out of n! permutations π from N to N, to satisfy
min(π(A)) = min (π(B))
*) π(S) is the set of integers obtained by applying permutation π to each element of S.
If min(π(A)) =min (π(B)), say y = π(x) is the common minimum.
Since the permutation π is a 1-to-1 mapping of N,
x ∈ A∩B
∴ A∩B cannot be empty.
⇒ y = π(x)
= π(A∩B) is the minimum of π(A∪B) is the minimum of π(A) and π(B) are to be same.
You can think like
*) If the minimum of π(A) and π(B) are same [min π(A)] = min [π(B)]
then min(π(A∩B)) = min(π(A∪B))
∴ Total number is given by n! |A∩B|/|A∪B|
*) Finally
Considering all possible permutations, the fraction of them that meet this condition |π(A∩B)| / |π(A∪B)|
[The probability of single permutation].
Ex: N = {1, 2, 3, 4} A = {1, 3} B = {1, 2, 4}

Since π is one to one mapping
|π(A∩B)| = |A∩B|
∴ π(A) = {1, 2}
π(B) = {1, 4, 3}
π(A∩B) = {1}
π(A∪B) = {1, 2, 3, 4}
4! × 1/4 = 6

Put g(i) = i+1

S = 1 + 2x + 3x² + 4x³ + .....
Sx = 1x + 2x² + 3x³ + 4x⁴ + ......
S - Sx = 1 + x + x² + x³ + .....
[Sum of infinite series in GP with ratio < 1 is a/1-r]
S - Sx = 1/(1-x)
S(1-x) = 1/(1-x)
S = 1/(1-x)²

No. of letters from A-Z is = 26
Each is printed twice the no. of letters = 26×2 = 52
If Mala has k colours, she can have k pairs of same colours.
She also can have ^kC₂ different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+^kC₂
k+^kC₂ ≥ 26
k+k(k-1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7

The possibilities to attend party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3ⁿ

Since we want at least k balls in each bag, so first we put kn balls into bags, k balls in each bag. Now we are left with m - kn balls, and we have to put them into n bags such that each bag may receive 0 or more balls. So applying theorem 2 of stars and bars with m - nk stars and n bars, we get number of ways to be

. So option (B) is correct

No. of cards = 52
No. of suits = 4(P)
Apply pigeon hole principal.
Then number of pigeons= n
floor [(n-1)/P] + 1 = 3
floor [(n-1)/P] = 2
floor [(n-1)] =8
floor (n) = 8 + 1
n ≥ 9
Minimum no. of cards, n = 9

Since there are n zeroes, so
XOXOXOXOXOXOX
n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are,
ⁿ⁺¹C_k

Formula for distributing n identical objects into r persons is,
^n+r-1C_r-1
So for 10 roses,
^10+2-1C_2-1 = ¹¹C₁ = 11
For 15 sunflowers,
^15+2-1C_2-1 = ¹⁶C₁ = 16
For 15 daffodils,
^15+2-1C_2-1 = ¹⁶C₁ = 16
∴ The final answer is,
11×16×16 = 2816

If the last digit is 0 then

If last digit is (2, 4, 6, 8)

Total possibilities = 504 + 1792 = 2296

The no. of ways,