Question 1

To fill 8 vacancies there are 15 candidates of which 5 are from ST. If 3 of the vacancies are reserved for ST candidates while the rest are open to all, Find the number of ways in which the selection can be done ?

Accepted Answer

ST candidates vacancies can be filled by C 3 5 ways = 10 Remaining vacancies are 5 that are to be filled by 12 => C 5 12 = (12x11x10x9x8)/(5x4x3x2x1) = 792 Total number of filling the vacancies = 10 x 792 = 7920

Question 2

The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is

Accepted Answer

The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in 7 P 6 ways. 7 teachers can sit in (7-1)! ways. Required no.of ways is = 7 P 6 . 6 ! = 7 ! . 6 !

Question 3

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

Accepted Answer

Number of ways of choosing 2 black pens from 5 black pens in 5 C 2 ways. Number of ways of choosing 2 white pens from 3 white pens in 3 C 2 ways. Number of ways of choosing 2 red pens from 4 red pens in 4 C 2 ways. By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

Question 4

On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students?

Accepted Answer

Given total number of students in the class = 21 So each student will have 20 greeting cards to be send or receive (21 - 1(himself)) Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

Question 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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

Permutation and Combination Questions