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This mock test of Combinatory MCQ - 1 for Computer Science Engineering (CSE) helps you for every Computer Science Engineering (CSE) entrance exam.
This contains 20 Multiple Choice Questions for Computer Science Engineering (CSE) Combinatory MCQ - 1 (mcq) to study with solutions a complete question bank.
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*Answer can only contain numeric values

QUESTION: 1

If the ordinary generating function of a sequence then a_{3} - a_{0} is equal to ___________ .

Solution:

a_{0} is the first term in the expansion of above series and a_{3} is the fourth term (or) coefficient of z^{3}

⇒ a_{3} − a_{0} = 16 − 1 = 15

*Answer can only contain numeric values

QUESTION: 2

The coefficient of x^{12} in (x^{3} + x^{4} + x^{5} + x^{6} + ....) is ___________.

Solution:

we will get x^{12} as

QUESTION: 3

The coefficient of x^{3} in the expansion of (1 + x)^{3 }(2 + x^{2})^{10} is.

Solution:

(1 + x)^{3} = (1 + x^{3} + 3x + 3x^{2})

As here we need to multiply last term of second expansion with first term of first coefficient (x^{3}) and 3x with x^{2} in the second expansion.

QUESTION: 4

We need to choose a team of 11 from a pool of 15 players and also select a captain. The number of different ways this can be done is

Solution:

Number of ways selecting a captain from 15 players =

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

QUESTION: 5

In how many different ways can elements be picked from a set of elements if

(i) Repetition is not allowed and the order of picking matters?

(ii) Repetition is allowed and the order of picking does not matter?

Solution:

(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

QUESTION: 6

There are kingdoms and champions. Each kingdom gets champions. The number of ways in which this can be done is:

Solution:

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) .

QUESTION: 7

The rules for the University of Bombay five-a-side cricket competition specify that the members of each team must have birthdays in the same month. What is the minimum number of mathematics students needed to be enrolled in the department to guarantee that they can raise a team of students?

Solution:

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.

QUESTION: 8

A 1 x 1 chessboard has one (1) square, a 2 x 2 chessboard has (5) squares. Continuing along this fashion, what is the number of squares on the (regular) 8 x 8 chessboard?

Solution:

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

QUESTION: 9

There is a set of 2n people: male and female. A good party is one with equal number of males and females (including the one where none are invited). The total number of good parties is.

Solution:

There are n men and n women

Now we can select 1 woman from n women in ^{n}C_{1}

With that 1 man can select ^{n}C_{1} ways

So, by 1 woman and 1 man we can get ^{n}C_{1} * ^{n}C_{1} ways...................i

Similarly , Now we can select 2 woman from n women in ^{n}C_{2}

With that 2 man can select ^{n}C_{2} ways

So, by 2 woman and 2 man we can get ^{n}C_{2} * ^{n}C_{2} ways.........................ii .........................................................

Now, by n woman and n man we can get ^{n}C_{n} * ^{n}C_{n} ways........................iii

So, by adding these equation we get

^{n}C_{0}.^{n}C_{0} + ^{n}C_{1} * ^{n}C_{1}+ ^{n}C_{2} * ^{n}C_{2} + ^{n}C_{3} * ^{n}C_{3}+.......................^{n}C_{n} * ^{n}C_{n} =(^{2n}C_{n})

Ans will be (E)

QUESTION: 10

Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at then it can move to either

Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)?

Solution:

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} =

*Answer can only contain numeric values

QUESTION: 11

A palindrome is a sequence of digits which reads the same backward or forward. For example, 7447, 1001 are palindromes, but 7455, 1201 are not palindromes. How many 8 digit prime palindromes are there?

Solution:

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).

QUESTION: 12

For each positive intefer n consider the set S_{n} defined as follows: S_{1} = {1}, S_{2} = {2, 3}, S_{3} = {4, 5, 6}, ... and, in general, S_{n + 1} consists of consecutinve integers the smallest of which is one more than the largest integer in S_{n} . Ten the sum of all the integers in S_{21} equals

Solution:

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 S_{21} using Arithmetic progression formula.

Let Sum(n) denote sum of natural numbers uptil n:

S_{1} starts with = 1

S_{2} starts with = Sum(1) + 1= 2

S_{3} starts with = Sum(2) + 1 = (1+2) + 1 = 4

Similarly S_{21} 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

QUESTION: 13

For the inter-hostel six-a-side football tournament, a team of 6 players is to be chosen from 11 players consisting of 5 forwards, 4 defenders and 2 goalkeepers. The team must include at least 2 forwards, at least 2 defenders and at least 1 goalkeeper. Find the number of different ways in which the team can be chosen.

Solution:

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.

QUESTION: 14

How many substrings (of all lengths inclusive) can be formed from a character string of length n ? Assume all characters to be distinct, prove your answer.

Solution:

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]

QUESTION: 15

How many distinct ways are there to split 50 identical coins among three people so that each person gets at least 5 coins?

Solution:

Distinct ways are there to split 50 identical coins among three people so that each person gets at least 5 coins

x_{1}+5+x_{2}+5+x_{3}+5 = 50

x_{1}+x_{2}+x_{3} = 35

Solving Non integral solution n = 35 ,r = 3

^{n+r-1} C _{r-1} = ^{35+3-1} C _{3-1} = ^{37}C_{2}

Hence E is Answer

QUESTION: 16

How many disctict words can be formed by permuting the letters of the word ABRACADABRA?

Solution:

QUESTION: 17

In a tournament with 7 teams, each team plays one match with every other team. For each match, the team earns two points if it wins, one point if it ties, and no points if it loses. At the end of all matches, the teams are ordered in the descending order of their total points (the order among the teams with the same total are determined by a whimsical tournament referee). The first three teams in this ordering are then chosen to play in the next round. What is the minimum total number of points a team must earn in order to be guaranteed a place in the next round?

Solution:

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

QUESTION: 18

The number of permutation of {1,2,3,4,5} that keep at least one integer fixed is.

Solution:

It is based on inclusion exclusion principle:

Total arrangements - derangements

5! - [5! - 5C_{1}(4!) + 5C_{2}(3!) - 5C_{3}(2!) + 5C_{4}(1!) - 5C_{5}(0!)]

5C_{1}(4!) - 5C_{2}(3!) + 5C_{3}(2!) - 5C_{4}(1!) + 5C_{5}(0!) = 120 - 60 + 20 - 5 + 1 = 76

QUESTION: 19

A club with x members is organized into tour committees such that

(a) each member is in exactly two committees,

(b) any two committees have exactly one member in common.

Then x has

Solution:

QUESTION: 20

A subset S of set of numbers {2,3,4,5,6,7,8,9,10} is said to be good if has exactly 4 elements and their gcd=1, Then number of good subset is

Solution:

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

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