1 Crore+ students have signed up on EduRev. Have you? 
In how many ways can three person, each throwing a single die once, make a score of 11
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
Match the pairs in the following questions by writing the corresponding letters only.
(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 twoelement 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.
It is required to divide the members of a club into disjoint teams of 2 members each. The teams are not labelled. The number of ways in which this can be done is:
2n member to be n teams with 2 member each and teams are unordered so we can exchange n team member among them.
In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
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 =
How many sub strings of different lengths (nonzero) can be found formed from a character string of length n ?
assuming an string of length n provided all alphabets are distinct..
no of strings of length 1 = n
no of strings of length 2 = n1
no of strings of length 3 = n2 .
.
.
no of string of length n = 1
total = n + (n 1) + (n  2) + (n  2) + ..... + 1
= n(n+1)/2
Find the number of binary strings w of length 2n with an equal number of 1's and 0's and the property that every prefix of w has at least as many 0's as 1's.
Answer to a is which is the Catalan number.
This is also equal to the number of possible combinations of balanced parenthesizes.
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
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that
i. each is sorted in ascending order,
ii. B has 5 and C has 3 elements, and
iii. the result of merging B and C gives A
select any 3 elements from given 8 elements  ^{8}C_{3}
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wifeneed not be accompanied by her husband. The number of different gatherings possible at the party is
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 = 3^{n}
m identical balls are to be placed in n distinct bags. You are given that m > kn, where is a natural number > 1. In howmany ways can the balls be placed in the bags if each bag must contain at least balls?
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 mk*n balls remaining.
We can use balls & sticks method now !
n bags= n variables, they need to equal to mk*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, n1 ) = C(m  k*n + n  1, m k*n )
Mala has the colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?
This question is slightly ambigous. So first let us understand what question is asking. So in a book, we have letters AZ 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.
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
How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)?
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 =
The number of binary strings of zeros and ones in which no two ones are adjacent is
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+1}C_{k}
Two girls have picked 10 roses, 15 sunflowers and 15 daffodils. What is the number of ways they can divide the flowers among themselves?
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
A pennant is a sequence of numbers, each number being 1 or 2. An pennant is a sequence of numbers with sum equal to n For example, (1,1,2) is a 4pennant. The set of all possible 1pennants is (1), the set of all possible 2pennants is (2), (1,1) and the set of all 3pennants is (2,1), (1,1,1,), (1,2). Note that the pennant (1,2) is not the same as thepennant (2,1). The number of 10pennants is________
Let we denote number of npennants by f(n), so f(10) is number of 10pennants.
So all 10pennants can be formed by 8pennants and 9pennants, 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
The number of 4 digit numbers having their digits in nondecreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is ________.
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 nondecreasing 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
The number of substrings (of all lengths inclusive) that can be formed from a character string of length n is
no. of substrings of length n is 1
no. of substrings of length n1 is 2
no. of substrings of length n2 is 3
so n(n+1)/2
In how many ways can b blue balls and r red balls be distributed in n distinct boxes?
r red balls can be distributed into n distinct boxes in C(n+r1,r) = .(n+r1)! / (n1)! r!
b blue balls can be distributed in C(n+b1,b) = (n+b1)! / (n1)! b!
By product rule total ways are (n+b1)! (n+r1)! / (n1)! b! (n1)! r!
SO THE ANSWER IS A.
In how many ways can we distribute 5 distinct balls, B_{1}, B_{2}, ..., B_{5} in 5 distinct cells, C_{1}, C_{2}, ...., C_{5} such that Ball Bi is not in cell and each cell contains exactly one ball?
Use Derangement concept D5= 44
so answer is A
A line L in a circuit is said to have a stuckat0 fault if the line permanently has a logic value 0. Similarly a line L in a circuit is said to have a stuckat1 fault if the line permanently has a logic value 1. A circuit is said to have a multiple stuckat fault if one or more lines have stuck at faults. The total number of distinct multiple stuckat faults possible in a circuit with N lines is
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)
Consider the recurrence relation a_{1} = 8, a_{n} = 6n^{2} + 2n + a_{n1}. Let a_{99} = K x 10^{4} . The value of K is __________.
a_{n} = 6n^{2} + 2n + a_{n−1}
= 6n^{2} + 2n + 6(n − 1)^{2} + 2(n − 1) + a_{n−2}
= 6n^{2} + 2n + 6(n − 1)^{2} + 2(n − 1) + 6(n − 2)^{2} + 2(n − 2)+. . . . . . +a_{1}
= 6n^{2} + 2n + 6(n − 1)^{2} + 2(n − 1) + 6(n − 2)^{2} + 2(n − 2)+. . . . . . +6.1^{2} + 2.1
= 6(n^{2} + (n − 1)^{2} +. . . +2^{2} + 1^{2}) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a_{99} = 2 x 99 x (99+1)^{2 }= 198 x 10^{4}
Let a_{n} be the number of bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for a_{n}?
Alternatively, we can get a string in a_{n} by appending "0" to any string in a_{n1} as well as by appending "01" to any string in a_{n2} and the two cases are mutually exclusive (no common strings) as well as exhaustive (covers all cases).
Consider the sequence S_{0},S_{1},S_{2},.... defined as follows: S_{0} = 0, S_{1} = 1 and S_{n} = 2S_{n1} + S_{n2} for n > 2. Which of the following statements is FALSE?
S_{n} = 2S_{n−1} + S_{n−2}
Characterstic polynomial for this recurrence is x^{2} = 2x + 1
x^{2} − 2x − 1 = 0
The solution to the recurrence relation is of the form :
Consider the sequencedefined by the recurrence relation where c >0.
Suppose there exists a nonempty, open interval (a,b) such that for all satisfying a < x_{0} < b , the sequence converges to a limit. The sequence converges to the value?
Since it converges, we can write:
x = cx^{2} − 2
or
cx^{2} − x − 2 = 0
Solving for x:
So both A) and B) can be the values.
Let H_{1}, H_{2}, H_{3}, ... be harmonic numbers. Then, for can be expressed as
The n^{th}Harmonic Number is defined as the summation of the reciprocals of all numbers from to .
3 videos7 docs100 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
3 videos7 docs100 tests









