Question 1

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 2

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 3

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 4

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 5

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.

Question 6

How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?

Accepted Answer

Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are 5 x 4 x 3 x 2 x 1 = 120.

Question 7

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Accepted Answer

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). Required number of ways = 6 C 1 * 4 C 3 + 6 C 2 * 4 C 2 + 6 C 3 * 4 C 1 + 6 C 4 = 6 C 1 * 4 C 1 + 6 C 2 * 4 C 2 + 6 C 3 * 4 C 1 + 6 C 2 = 209.

Question 8

How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never beings separated?

Accepted Answer

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways. But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

Question 9

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

Accepted Answer

There are 8 students and the maximum capacity of the cars together is 9. We may divide the 8 students as follows Case I: 5 students in the first car and 3 in the second Case II: 4 students in the first car and 4 in the second Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways. Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways. Therefore, the total number of ways in which 8 students can travel is: 8 C 3 + 8 C 4 =56 + 70= 126

Question 10

From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee…

Accepted Answer

We first count the number of committee in which (i). Mr. Y is a member (ii). the ones in which he is not Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member. Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join). We can choose 1 more in5+ 2 C 1 =7 ways. Case (ii): If Mr. Y is not a member then we left with (6+4-1) people. we can select 3 from 9 in 9 C 3 =84 ways. Thus, total number of ways is 7+84= 91 ways.

Question 11

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

Accepted Answer

The given word HAPPYHOLI has 9 letters These 9 letters can e arranged in 9! ways. But here in the given word letters H & P are repeated twice each Therefore, Number of ways these 9 letters can be arranged is 9 ! 2 ! x 2 ! = 9 x 8 x 7 x 6 x 5 x 4 x 3 2 = 90 , 720 ways .

Question 12

The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements?

Accepted Answer

Given Word is PROMISE. Number of letters in the word PROMISE = 7 Number of ways 7 letters can be arranged = 7! ways Number of Vowels in word PROMISE = 3 (O, I, E) Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways Now, the number of ways of arrangements so that three vowels should not come together = 7! - (5! x 3!) ways = 5040 - 720 = 4320.

Question 13

In how many ways can letter of the word RAILINGS arrange so that R and S always come together?

Accepted Answer

The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2! But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways.

Question 14

In a bag, there are 8 red, 7 blue and 6 green flowers. One of the flower is picked up randomly. What is the probability that it is neither red nor green ? A) 13 B)821 C)621 D)2021

Accepted Answer

Total number of flowers = (8+7+6) = 21. Let E = event that the flower drawn is neither red nor green. = event taht the flower drawn is blue. --> n(E)= 7 --> P(E)= 7 21 = 1 3

Question 15

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ?

Accepted Answer

We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys). Required number of ways = ( C 1 6 × C 3 4 ) + C 2 6 × C 2 4 + ( C 3 6 × C 1 4 ) + ( C 4 6 ) = (24+90+80+15) = 209.

Question 16

In how many different ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together?

Accepted Answer

Given word is TRANSFORMER. Number of letters in the given word = 11 (3 R's) Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is 10! x 2!/3! = 3628800 x 2/6 = 1209600

Question 17

What is the value of 100P2 ?

Accepted Answer

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged. That can be done as, 100 P 2 = 100!/(100 - 2)! = 100 x 99 x 98!/98! = 100 x 99 = 9900.

Question 18

Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

Accepted Answer

We have to arrange 6 books. The number of permutations of n objects is n! = n. (n – 1) . (n – 2) ... 2.1 Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720

Permutation and Combination Questions