GATE Computer Science Engineering(CSE) 2027 Test: Combinatory- 2 Free Online

The sum 11 can be broken as 6 + 5
Let us name the players as A, B and C.
Case 1: Fix 6 and break 5

Suppose A throws 6. Players B and C can throw dice in following four ways to make sum 5
(1,4), (2,3), (3,2) and (4,1)
As any of the three players can throw value 6, so there are total 3 x 4 = 12

Case 2: Fix 5 and break 6

Suppose A throws 5. Players B and C can throw dice in following five ways to make sum 6
(1,5), (2,4), (3,3), (4,2), (5,1)
As any of the three players can throw value 5, so there are total 3 x 5 = 15
Adding case 1 and case 2 there are total  12 + 15 = 27 ways

(A) - S Catalyn no
(B) - R. Choosing n locations out of 2n to place 0. Remaining automatically become 1.
(C) -P An even permutation is a permutation obtainable from an even number of two-element swaps, For a set of elements and n > 2, there are  n! / 2 even permutations.
(D) -> Q 
Length = 6n, as it is palindrome, we need to only consider half part.
Total Length to consider 3n (Remaining 3n will be revese of this 3n)
now Choosing n 0's out of 3n. So Q is correct for D.

2n member to be n teams with 2 member each and teams are unordered so we can exchange n team member among them.

Take AAU together and treat it like 1 entity. Now arrange 
Then, the AAU can be arranged in ways because A has been repeated twice.
So, total arrangements = 

assuming an string of length n provided all alphabets are distinct..
no of strings of length 1 = n
no of strings of length 2 = n-1
no of strings of length 3 = n-2 .
.
.
no of string of length n = 1
total = n + (n -1) + (n - 2) + (n - 2) + ..... + 1

= n(n+1)/2

Answer to a is which is the Catalan number.
This is also equal to the number of possible combinations of balanced parenthesizes. 

• If the number ends with a 0 then there are 9 choices for the first digit, 8 for the second and 7 for the third, which makes 1×9×8×7=504 possibilities.
• If the number is even ending with something else than 0 then there are 4 choices for the last digit, 8 choices for the first digit (no 0 nor the last digit), 8 for the second digit and 7 for the third digit, which makes 4×8×8×7=1792

​Together, this gives 2296 numbers with 4 distinct digits that are even. Note that this does not allow leading 0, as you see to want it based from the question

select any 3 elements from given 8 elements - ⁸C₃

Possible outcome for a couple:

1. only wife comes
2. both come
3. none come

Thus 3 possibilities for each couple, so 3 x 3 x 3 x ... n times = 3ⁿ

As there have to be atleast k balls in each bag, so firstly put k balls in each bag i.e k*n balls.
Then now we have total m-k*n balls remaining.
We can use balls & sticks method now !
n bags= n variables, they need to equal to m-k*n, no restrictions on how many balls in each bag !
x1 + x2 + ... + xn = m- k*n , x1,x2..xn >=0.
So when we solve it
We get
C(m - k*n + n - 1, n-1 ) = C(m - k*n + n - 1, m- k*n )

This question is slightly ambigous. So first let us understand what question is asking. So in a book, we have letters A-Z and each letter is printed twice, so there are 52 letters. Now we have to color each letter, so we need a pair of colors for that, because each letter is printed twice. Also in a pair, both colors can be same. Now condition is that a pair of colors can't be used more than once.
So suppose Mala has 3 colors : Red, Blue, Green. She can color as follows : 1:(Red,Red),  2:(Blue,Blue), 3:(Green,Green), 4: (Red,Blue), 5: (Red,Green),
6 : (Blue,Green).
Now we don't have more pairs of colors left, we have used all pairs, but could color only 6 letters out of 26. So question is to find minimum no. of colors, so that we could color all 26 letters.
So if Mala has k colors, she can have k pairs of same colors, thus coloring k letters, then kC2 other pairs of colors, thus coloring kC2 more letters. 

So total no. of letters colored 
So we want 

so option (C) is correct.

Say,  r = Move Right and 
u = Move Up
so using 10 combination of r and 10 combinations of u moves we get a solution.

Convert the graphical moves to text and one such solution we get =
{u, u, u, u, u, u, u, u, u, u, r, r, r, r, r, r, r, r, r, r} now all possible arrangements of them is given by =

first place n zeroes side by side _ 0 _ 0 _ 0 ... 0 _
k 1's can be placed in any of the (n+1) available gaps hence number of ways  = ⁿ⁺¹C_k

number of ways roses can be distributed - {(0, 10), (1, 9), (2, 8).....(10, 0)} - 11 ways
similarly sunflowers and daffodils can be distributed in 16 ways each
total number of ways 11 x 16 x 16 = 2816

Let we denote number of n-pennants by f(n), so f(10) is number of 10-pennants.

So all 10-pennants can be formed by 8-pennants and 9-pennants, and no other pennant (since we can add only 1 or 2 into a sequence)
So f(10) = f(9) + f(8)
This is in fact a fibonacci sequence, in which F(1) = 1, f(2) = 2, so this sequence becomes
1, 2, 3, 5, 8, 13, 21, 34, 55, 89,..
So f(10) = 89.

Numbers could be any one of
(1,1,1,1,1,1,1,1,1,1),(1,1,1,1,1,1,1,1,2),(1,1,1,1,1,1,2,2),(1,1,1,1,2,2,2),(1,1,2,2,2,2),(2,2,2,2,2)
So, the number of 10 penants =1+  9!/8! + 8!/6!2!  + 7!/4!3! + 6!/2!4! +1 =89

Dynamic programming Approach

Here Starting 1 means numbers starting with 1. And cell is for number of numbers starting with  and having  digits.
We can have the relation
 

as per the non-decreasing condition given in the question. So, our answer will be c(1, 4) + c(2, 4) + c(3, 4) = 1 + 4 + 10 = 15

Brute force
3 3 3 3
2 2 2 2
2 2 2 3 
2 2 3 3
2 3 3 3
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 3 3 3

you can arrive at a solution by constructing a graph for each starting digit. for example root 3 means - starting with 3 it can have 3 child 1,2,3 and the construction goes
3 can three children 1, 2,3
2 can have two children 1, 2
1 can have only 1 as child. Graph need to be done till three levels and finally count total number of leaves

no. of substrings of length n is 1
no. of substrings of length n-1 is 2
no. of substrings of length n-2 is 3
so n(n+1)/2

r red balls can be distributed into n distinct boxes in C(n+r-1,r) = .(n+r-1)! / (n-1)! r!
b blue balls can be distributed in C(n+b-1,b) = (n+b-1)! / (n-1)! b!
By product rule total ways are (n+b-1)! (n+r-1)! / (n-1)! b! (n-1)! r!
SO THE ANSWER IS  A.

Use Derangement concept D5= 44
so answer is A

Answer should be 3^N-1.

This is because the total possible combinations (i.e a line may either be at fault (in 2 ways i.e stuck at fault 0 or 1) or it may not be , so there are only 3 possibilities for a line) is 3^N. In only one combination the circuit will have all lines to be correct (i.e not at fault.) Hence 3^N-1. (as it has been said that circuit is said to have multiple stuck up fault if one or more line is at fault)

a_n = 6n² + 2n + a_n−1
= 6n² + 2n + 6(n − 1)² + 2(n − 1) + a_n−2
= 6n² + 2n + 6(n − 1)² + 2(n − 1) + 6(n − 2)² + 2(n − 2)+. . . . . . +a₁
= 6n² + 2n + 6(n − 1)² + 2(n − 1) + 6(n − 2)² + 2(n − 2)+. . . . . . +6.1² + 2.1
= 6(n² + (n − 1)² +. . . +2² + 1²) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a₉₉ = 2 x 99 x (99+1)² = 198 x 10⁴

Alternatively, we can get a string in  a_n by appending "0" to any string in  a_n-1 as well as by appending "01" to any string in a_n-2 and the two cases are mutually exclusive (no common strings) as well as exhaustive (covers all cases). 

S_n = 2S_n−1 + S_n−2

Characterstic polynomial for this recurrence is x² = 2x + 1

x² − 2x − 1 = 0 
The solution to the recurrence relation is of the form  :

Since it converges, we can write:
x = cx² − 2
or 
cx² − x − 2 = 0
Solving for x:

So both A) and B) can be the values.

The  n^thHarmonic Number  is defined as the summation of the reciprocals of all numbers from  to .