Question 1

From a bunch of flowers having 16 red roses and 14 white roses, four flowers have to be selected. In how many different ways can they be selected such that at least one red rose is selected?

Accepted Answer

Given total 16 Red roses and 14 White roses = 30 roses Four flowers have to be selected from 30 i.e, C 4 30 = 27405 Ways Now, atleast one Red rose is selected i.e, 27405(total) - 1(all four are white roses) = 27404 ways.

Question 2

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king?

Accepted Answer

Number of cards in a pack of cards = 52 Number of black cards = 26 Number of king cards = 4 (2 Red, 2 Black) Required, the probability that if a card is drawn either card is black or a king = 26 52 + 4 52 = 30 52 = 15 26

Question 3

In how many different ways can the letters of the word 'RITUAL' be arranged?

Accepted Answer

The number of letters in the given word RITUAL = 6 Then, Required number of different ways can the letters of the word 'RITUAL' be arranged = 6! => 6 x 5 x 4 x 3 x 2 x 1 = 720

Question 4

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

Accepted Answer

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. (i) 1 lady out of 4 and 4 gentlemen out of 6 (ii) 2 ladies out of 4 and 3 gentlemen out of 6 (iii) 3 ladies out of 4 and 2 gentlemen out of 6 (iv) 4 ladies out of 4 and 1 gentlemen out of 6 In case I the number of ways = C 1 4 × C 4 6 = 4 x 15 = 60 In case II the number of ways = C 2 4 × C 3 6 = 6 x 20 = 120 In case III the number of ways = C 3 4 × C 2 6 = 4 x 15 = 60 In case IV the number of ways = C 4 4 × C 1 6 = 1 x 6 = 6 Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

Question 5

In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends?

Accepted Answer

DESIGN = 6 letters No consonants appear at either of the two ends. = 2 x 4 P 4 = 2 x 4 x 3 x 2 x 1= 48

Question 6

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

Accepted Answer

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5. Number of ways of arranging the vowels = 3 P 3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3 P 3 = 3! = 6. Total number of ways = (6 x 6) = 36.

Question 7

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

Accepted Answer

The first letter from the right can be chosen in 26 ways because there are 26 alphabets. Having chosen this, the second letter can be chosen in 26 ways The first two letters can chosen in 26 x 26 = 676 ways Having chosen the first two letters, the third letter can be chosen in 26 ways. All the three letters can be chosen in 676 x 26 =17576 ways. It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

Question 8

How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters ?

Accepted Answer

Each of the 10 letters can be posted in any of the 5 boxes. So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters. i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

Question 9

How many words can be formed with or without meaning by using three letters out of k, l, m, n, o without repetition of alphabets.

Accepted Answer

Given letters are k, l, m, n, o = 5 number of letters to be in the words = 3 Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5 P 3 ways . ⇒ 5 P 3 = 5 x 4 x 3 = 60 words .

Question 10

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

Accepted Answer

15! - 14! - 13! = (15 × 14 × 13!) - (14 × 13!) - (13!) = 13! (15 × 14 - 14 - 1) = 13! (15 × 14 - 15) = 13! x 15 (14 - 1) = 15 × 13 × 13!

Question 11

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow ?

Accepted Answer

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35 Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105 Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35 Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21

Question 12

There are 11 True or False questions. How many ways can these be answered ?

Accepted Answer

Given 11 questions of type True or False Then, Each of these questions can be answered in 2 ways (True or false) Therefore, no. of ways of answering 11 questions = 2 11 = 2048 ways.

Question 13

The ratio of the members of blue to red balls in abag is constant. When there were 44 red balls, the number of blue balls was 36. If the number of blue balls is 54, how many red balls will be in the bag?

Accepted Answer

66

Question 14

How many more words can be formed by using the letters of the given word 'CREATIVITY'?

Accepted Answer

The number of letters in the given word CREATIVITY = 10 Here T & I letters are repeated => Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200

Question 15

A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ? A) 11/91 B) 4/11 C) 1/11 D) 90/91

Accepted Answer

Let S be the sample space. Then, n(s) = number of ways of drawing 4 pearls out of 14 = C 4 14 ways = 14 × 13 × 12 × 11 4 × 3 × 2 × 1 = 1001 Let E be the event of drawing 4 pearls of the same colour. Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5) ⇒ C 1 5 + C 4 4 + C 1 5 = 5+1+5 =11 P(E) = n ( E ) n ( S ) = 11 1001 = 1 91 Required probability = 1 - 1 91 = 90 91

Question 16

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?

Accepted Answer

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5. So, there is one way of doing it. Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

Question 17

Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

Accepted Answer

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) =( C 3 7 × C 2 4 ) = 210. Number of groups, each having 3 consonants and 2 vowels =210 Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 120. Required number of words = (210 x 120) = 25200.

Question 18

A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ?

Accepted Answer

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways. The captain can be selected from amongst the remaining 3 players in 3 ways. Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways. (or) Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 28x6 = 168 ways.

Question 19

36 identical books must be arranged in rows with the same number of books in each row. Each row must contain at least three books and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are …

Accepted Answer

The following arrangements satisfy all 3 conditions. Arrangement 1: 3 books in a row; 12 rows. Arrangement 2: 4 books in a row; 9 rows. Arrangement 3: 6 books in a row; 6 rows. Arrangement 4: 9 books in a row; 4 rows. Arrangement 5: 12 books in a row; 3 rows. Therefore, the possible arrangements are 5.

Permutation and Combination Questions