Question 1 1 / -0
In how many ways can a garland be made from exactly 10 flowers?
Question 2 1 / -0
The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is
Solution
You have two different kinds of such three-digit even numbers.First is 5 at the hundred'splace and second 5 is not at the hundred'splace
In first case no is of the form 57x, where x is the unit's digit ,which can be 0,2,4,6,8 which is just 5 possibilities.Hence the no of possibilities in this case is 1x1x5=5
In second case the hundred's digit can be 1,2,3,,4,,6,7,8,or,9 which is 8 ways and the ten's digit can be any of the 9 numbers and unit digit can be any of the 5 even numbers .Therfore the no: of ways will be 8×9×5=360 So total we have 360 + 5 = 365 possibilities
Question 3 1 / -0
Out of 9 consonants and 5 vowels, how many words of 3 consonants and 3 vowels can be formed?
Question 4 1 / -0
In how many ways can a mixed doubles tennis game be arranged from a group of 10 players consisting of 6 men and 4 women
Solution
A team of 4 players are to be selected.
2 out of 6 men can be done in 6 C2 ways. 2 out of 4 women can be done in 4 C2 ways.
So the number of ways to select 4 players is 6 C2 x 4 C2 = 90.
Now we can arrange these people to form mixed doubles.
If M1 ,M2 ,W1 ,W2 , are the 4 members selected then one team can be chosen as (M1 ,W1 )or(M1 ,W2 ) in 2 different ways
Therefore the required number of arrangements= 90x2 = 180.
Question 5 1 / -0
If no husband and wife play in the same game, then the number of ways in which a mixed double game can be arranged from among nine married couples is
Question 6 1 / -0
A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
Solution
Let the friends be A,B,C,D,E,F,G,H,I , J and assume A and B will not battend together
Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members
Then no of ways is 8 C6 = 28 ways.
Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and one from A and B
Then the no of ways = 2 C1 x 8 C5 =112 ways.
Therefore total number of ways is 28+ 112 = 140 ways,
Question 7 1 / -0
The maximum number of points at which 4 circles and 4 straight lines can intersect is
Question 8 1 / -0
Out of his 20 friends, a person wishes to make up as many different groups as he can such that each group consists of the same number of persons. The number of friends he should invite at a time is
Question 9 1 / -0
What is the number of arrangements that can be made using all the letters of the word 'LAUGH', keeping the vowels adjacent?
Question 10 1 / -0
What will be the sum of all the numbers greater than 1000 formed by using the digits 1, 3, 5 and 7, such that no digit is being repeated in any number?
Question 11 1 / -0
Everybody in a room shakes hand with every other person. The total number of handshakes is 66. Find the total number of persons in the room.
Question 12 1 / -0
The sides AB, BC and CA of a triangle ABC have 3, 4 and 5 points lying on them, respectively. The number of triangles that can be constructed using these points as vertices is
Question 13 1 / -0
In a function, 10 persons including A, B and C have to perform. If A has to perform before B and B has to perform before C, how many different ways are there to arrange the order of their performances?
Question 14 1 / -0
On the occasion of Diwali, each student of a class sends greeting cards to every other. If there are 20 students in the class, then the total number of greeting cards exchanged by the students is
Question 15 1 / -0
A basket contains 12 red, 10 black and 5 pink balls. 5 balls are drawn at random. The number of ways of drawing at least 2 black balls is
Question 16 1 / -0
H.C.F. of n !, (n + 1) ! and (n + 2) ! is
Question 17 1 / -0
The number of ways in which p + q things can be divided into two groups, containing p and q things respectively is
Question 18 1 / -0
The number of all possible selections which a student can make for answering one or more questions out of eight given questions in a paper, where each question has one alternate option, is
Question 19 1 / -0
The number of significant numbers which can be formed by using any number of the digits 0, 1, 2, 3, 4
Solution
Case 1: for 5 digit numbers Ten Tens thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways and the thousands place in 1 way. Hence the total number of ways is 4x1x4x2x3= 96
Case 2: for 4 digit numbers thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways . Hence the total number of ways is 4x4x2x3= 96.
Case 3: for 3digit numbers hundreds place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways Hence the total number of ways is 4x4x3= 48
Case 4: for 2 digit numbers Tens place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied. Hence the total number of ways is 4x4=16.
Case 5: For single digit in 4 ways .
Hence 96 + 96 + 48+ 16+4 = 260
Question 20 1 / -0
How many words can be formed by taking four different letters of the word MATHEMATICS?
Question 21 1 / -0
A coin is tossed n times, the number of all the possible outcomes is
Solution
There can either be a heads or tails, therefore for every toss, the possible outcomes are 2. hence for n number of toss the possibilities are 2n.
Question 22 1 / -0
If n C4, n C5 and n C6 are in A.P, then n is equal to
Question 23 1 / -0
A cricket team of eleven is to be chosen from among 8 batsmen, 6 bowlers and 2 wicket-keepers. In how many ways can the team be chosen, if there must be at least four batsmen, at least four bowlers and exactly one wicket-keeper?
Question 24 1 / -0
Eight chairs are numbered from 1 to 8. 2 women and 3 men wish to occupy one chair each. First, the women choose the chairs from amongst the chairs marked from 1 to 4 and then, the men select the chairs from amongst the remaining. The number of possible arrangements is
Question 25 1 / -0
The number of ways in which 5 beads of different colours form a necklace is
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