GATE Computer Science Engineering(CSE) 2027 Test: Combinatory- 1 Free Online

a₀ is the first term in the expansion of above series and  a₃ is the fourth term (or) coefficient of z³

⇒ a₃ − a₀ = 16 − 1 = 15

we will get x¹² as 

(1 + x)³ = (1 + x³ + 3x + 3x²)

As here we need to multiply last term of second expansion with first term of first coefficient (x³) and 3x with x² in the second expansion.

Number of ways selecting a captain from 15 players = 

Number of ways selecting remaining team members from remaining 14 players = 

(i) Repetition is not allowed and the order of picking matters =
= r arrangement with no repetition 
(ii) Repetition is allowed and the order of picking does not matter =  
= combination with unlimited repetition

Option A is correct.
We have n Kingdoms as k1 , k2 , ... , kn.
Firstly we can select 2 champions from 2n champions and assign to k1 = ways(Say w1)

Then we can select next 2 champions and assign to k2 =  ways(Say w2)
and so on..

For last kingdom , we have 2 champions left = 
ways  (say w_n)

Total ways for assigning 2n champions to n kingdoms 
=

= (2n)! / 2ⁿ So, Option A  (Ans) . 

There are 12 months and we have to get 5 people having birthdays in the same month in order to form a team . we can apply the pigeon hole principal :

[N/12] = 5

On solving we get N=49 .
Hence answer is D.

no of squares on chessboard of n*n is equal to sum of squares of n terms
for 8*8 chessboard
= n(n+1)(2n+1)/6
=8*9*17/6
=204

There are n men and n women
Now we can select 1 woman from n women in ⁿC₁
With that 1 man can select  ⁿC₁ ways
So, by 1 woman and 1 man we can get   ⁿC₁ * ⁿC₁ ways...................i
Similarly , Now we can select 2 woman from n women in ⁿC₂
With that 2 man can select  ⁿC₂ ways
So, by 2 woman and 2 man we can get   ⁿC₂ * ⁿC₂ ways.........................ii .........................................................
Now, by n woman and n man we can get   ⁿC_n * ⁿC_n ways........................iii
So, by adding these equation we get
ⁿC₀.ⁿC₀ + ⁿC₁ * ⁿC₁+  ⁿC₂ * ⁿC₂ + ⁿC₃ * ⁿC₃+.......................ⁿC_n * ⁿC_n  =(²ⁿC_n)

Ans will be (E)

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 =

now we need to discard the segment move from
(4, 4) to 
(5, 5) :
to do that we first calculate how many solutions to our problem to reach (10, 10) involves that segment. We'll then subtract those solutions from the total number of solutions.
Number of solutions to reach from (0,0) to (4,4) = all possible arrangements of {r, r, r, r, u, u, u, u} = 
 
definitely we take the segment (4,4) to (5,4) = 1
now, Number of solutions to reach from (5,4) to (10,10) = all possible arrangements of {r, r, r, r, r, r, u, u, u, u, u} =

I think the answer should be 0, as any even digit palindrome(other than 11) cannot be prime. Even digit palindromes will always be divisible by 11(you can check the divisibility test by 11).

Every new set S_n+1 starts after n elements from the starting element of S_n . This means that we can find the starting number of S₂₁ using Arithmetic progression formula.
Let Sum(n) denote sum of natural numbers uptil n:
S₁ starts with = 1
S₂ starts with  = Sum(1) + 1= 2
S₃ starts with = Sum(2) + 1 = (1+2) + 1 = 4
Similarly S₂₁ starts with S(20) +1 = 
Now we need to find sum of 21 consecutive natural numbers starting from 211 Using A.P. sum formula where a= starting term , d= difference
 4641... Option D is correct

There are three ways to choose 6 Players.
1. 5C3*4C2*2C1=120
2. 5C2*4C2*2C2=60
3. 5C2*4C3*2C1=80
So total No of ways is 260.

Lets take an example . lets consider the given string is GATE.
so set of string of length 1 ={G,A,T,E} ; cardinality of set = 4
set of string of length 2 ={GA,AT,TE}
set of string of length 3={GAT,ATE}
set of strings of length 4 ={GATE}
and set of string of length 0 ={}
and we cant have any substring of length 5 as given string has only 4 length.
so total no of substrings are possible =0's length substring + 1lengths substrings + 2 length substrings +3 length substrings + 4 length substrings = 1+4+3+2+1
means for 1 length string to n length substrings . it will sum of the n natural no from 1 to n.
so 1+2+3+............... +n = n(n+1)/2
so total no substrings possible = 0 length strings + n(n+1)/2= 1+[n(n+1)/2]
so total no of substrings possible in n length string (All length inclusive) = 1+[n(n+1)/2]

Distinct ways are there to split 50 identical coins among three people so that each person gets at least 5 coins
x₁+5+x₂+5+x₃+5 = 50
x₁+x₂+x₃ = 35
Solving Non integral solution n = 35 ,r = 3
^n+r-1 C _r-1 = ^35+3-1 C _3-1 = ³⁷C₂

Hence E is Answer

Let the 7 Teams be A,B,C,D,E,F,G and so each team plays total 6 matches.
Suppose, Team A wins over E,F,G and draws with B,C,D hence total points scored by Team A = 9 points
Now, Team B wins over E,F,G  and draws with A,C,D hence total points scored by Team B = 9 points
Similarly, happens for next two teams C and D .
Hence, Finalized scores are => 

Given that the order among the teams with the same total are determined by a whimsical tournament referee.
So, He/She can order the top 3 teams like ABC,ABD ,BCD ,ACD .......
But, Question says " team must earn in order to be guaranteed a place in the next round "
Hence, Not to depend on that whimsical referee, the minimum total number of points a team must earn in order to be guaranteed a place in the next round = 9 + 1 = 10 points

It is based on inclusion exclusion principle:
Total arrangements - derangements
5! - [5! - 5C₁(4!) + 5C₂(3!) - 5C₃(2!) + 5C₄(1!) - 5C₅(0!)]
5C₁(4!) - 5C₂(3!) + 5C₃(2!) - 5C₄(1!) + 5C₅(0!) = 120 - 60 + 20 - 5 + 1 = 76 

Selecting 4 numbers from 9 is 9C4 = 126
We have to subtract all the cases where gcd of all numbers is 1 and this can only happen when all number should not be even. Which is 5C4 = 5
So answer will be 126 - 5 = 121