Question 1

Some children goto ice-cream shop. 9 flavours are available there. Each child takes a cone with two different flavours. No two children take same combination and they have taken all such possible combinations. How many children went to ice cream sho…

Accepted Answer

Given there are 9 flavours of ice creams. Each child takes the combination of two flavours and no two children will have the same combination This can be done by ways i.e children. Number of children= C 2 9 = 9 x 8 / 1x 2 = 36.

Question 2

How many three digit numbers 'abc' are formed where two of the three digits are same ?

Accepted Answer

Digits are 0,1,2,3,4,5,6,7,8,9. So no. of digits are 10 First all possible case => 9(0 excluded) x 10 x 10 = 900 Second no repetition allowed =>9 x 9 x 8 = 648 Third all digits are same => 9 (111,222,333,444,555,666,777,888,999) Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

Question 3

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is ?

Accepted Answer

We can initially arrange the six boys in 6! ways. Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways. Hence required number of ways = 6! x ⁷P₆

Question 4

From a group of 7 boys and 6 girls, five persons are to be selected to form a team, so that at least 3 girls are there in the team. In how many ways can it be done?

Accepted Answer

Given in the question that, there are 7 boys and 6 girls. Team members = 5 Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 6 C 3 x 7 C 2 + 6 C 4 x 7 C 1 + 6 C 5 = 6 x 5 x 4 3 x 2 x 1 x 7 x 6 2 x 1 + 6 x 5 x 4 x 3 4 x 3 x 2 x 1 x 7 + 6 x 5 x 4 x 3 x 2 5 x 4 x 3 x 2 x 1 = 420 + 105 + 6 = 531 .

Question 5

In a plane 8 points are colliner out of 12 points, then the number of triangles we get with those 12 points is

Accepted Answer

For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is 12 C 3 . But among them 8 points are collinear. If all these 8 points are different we get 8 C 3 triangles as they are collinear. In 12 C 3 triangles, we do not get 8 C 3 triangles Therefore, The number of triangles we get = 12 C 3 - 8 C 3 = 164

Question 6

There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairytales, novels and plays ?

Accepted Answer

There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways. Similarly, there are 5 novels.They can be arranged in 5! ways. And there are 3 plays.They can be arranged in 3! ways. So, by the counting principle all of them together can be arranged in 4!´5!´3! ways = 17280

Question 7

How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?

Accepted Answer

1 million distinct 3 digit initials are needed. Let the number of required alphabets in the language be ‘n’. Therefore, using ‘n’ alphabets we can form n * n * n = n 3 distinct 3 digit initials. Note distinct initials is different from initials where the digits are different. For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different. This n 3 different initials = 1 million i.e. n 3 = 10 6 (1 million = 10 6 ) => n = 10 2 = 100 Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

Question 8

A polygon 7 sides.How many diagonals can be formed?

Accepted Answer

7 C 2 - 7 = 14

Question 9

If there are 15 dots on a circle,how many triangles can be formed?

Accepted Answer

There are 15 dots in total,and to make a triangle we need to select any three of those dots. So, 15 C 3 = 455

Question 10

When six fairs coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads ?

Accepted Answer

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads. i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads. The number of outcomes in which 0 coins turn heads is 6 C 0 =1 The number of outcomes in which 1 coin turns head is = 6 C 1 =6 The number of outcomes in which 2 coins turn heads is 6 C 2 =15 The number of outcomes in which 3 coins turn heads is 6 C 3 =20 Therefore, total number of outcomes =1+6+15+20= 42 outcomes

Question 11

There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of th…

Accepted Answer

All the boxes contain distinct number of chocolates. For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on. The number of ways of giving 4 boxes to the 4 person is: 8 C 4 = 70

Question 12

If the letters of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word VERMA is :

Accepted Answer

The number of words beign with A is 4! The number of words beign with E is 4! The number of words beign with M is 4! The number of words beign with R is 4! Number of words beign with VA is 3! Words beign with VE are VEAMR VEARM VEMAR VEMRA VERAM VERMA Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108

Question 13

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed ?

Accepted Answer

The Word LOGARITHMS contains 10 letters. To find how many 4 letter words we can form from that = P 4 10 =10x9x8x7 = 5040.

Question 14

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

Accepted Answer

No. of letters in the word = 6 No. of 'E' repeated = 2 Total No. of arrangement = 6!/2! = 360

Question 15

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

Accepted Answer

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E Total = 13 letters But last letter must be N Hence, available places = 12 In that odd places = 1, 3, 5, 7, 9, 11 Owvels = 4 This can be done in 6P4 ways Remaining 7 letters can be arranged in 7!/3! x 2! ways Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

Question 16

A box contains 5 green, 4 yellow and 3 white marbles. Threemarbles are drawn at random. What is the probability thatthey are not of the same colour ?

Accepted Answer

n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220

Question 17

In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?

Accepted Answer

They have to be arranged in the following way : L T L T L T L T L The 5 lions should be arranged in the 5 places marked ‘L’. This can be done in 5! ways. The 4 tigers should be in the 4 places marked ‘T’. This can be done in 4! ways. Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.

Question 18

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Accepted Answer

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). Required number of ways= 7 C 3 * 6 C 2 + 7 C 4 * 6 C 1 + 7 C 5 = 756.

Question 19

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Accepted Answer

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black). Required number of ways= 3 C 1 * 6 C 2 + 3 C 2 * 6 C 1 + 3 C 3 = (45 + 18 + 1) =64

Question 20

A standard deck of playing cards has 13 spades. How many ways can these 13 spades be arranged?

Accepted Answer

The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial. 13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

Permutation and Combination Questions