Question 1

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 2

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 3

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 4

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 5

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

Question 6

A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible …

Accepted Answer

Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252

Question 7

There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when ea…

Accepted Answer

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities. The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier). For each value on the card the remainder can have 3 possible values. The total number of possible sequences is: 4*3^4

Question 8

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any…

Accepted Answer

Let the number of Rose plants be ‘a’. Let number of marigold plants be ‘b’. Let the number of Sunflower plants be ‘c’. 20a+5b+1c=1000; a+b+c=100 Solving the above two equations by eliminating c, 19a+4b=900 b = (900-19a)/4 b = 225 - 19a/4----------(1) b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2) Substituting (1) in (2), 0 < 225 - 19a/4 < 99 225 < -19a/4 < (99 -225) => 4 x 225 > 19a > 126 x 4 => 900/19 > a > 505 a is the integer between 47 and 27 ----------(3) From (1), it is clear, a should be multiple of 4. Hence possible values of a are (28,32,36,40,44) For a=28 and 32, a+b>100 For all other values of a, we get the desired solution: a=36,b=54,c=10 a=40,b=35,c=25 a=44,b=16,c=40 Three solutions are possible.

Question 9

In how many ways can 100 soldiers be divided into 4 squads of 10, 20, 30, 40 respectively?

Accepted Answer

100 C 10 + 90 C 20 + 70 C 30 + 40 C 40 = 100 ! 10 ! x 20 ! x 30 ! x 40 !

Question 10

What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition ?

Accepted Answer

We know that zero can't be in hundreds place. But let's assume that our number could start with zero. The formula to find sum of all numbers in a permutation is 111 x no of ways numbers can be formed for a number at given position x sum of all given digits No of 1 s depends on number of digits So,the answer us 111 x 20 x (0+1+2+3+4+5) = 33300 We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula. Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0. This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number. So the sum for this is 11 x 4 x (1+2+3+4+5). =660 Hope u understood why we use 4. Each number can be formed in 4x1 ways So, the final answer is 33300-660 = 32640

Question 11

There are 3 bags, in 1st there are 9 Mangoes, in 2nd 8 apples & in 3rd 6 bananas. There are how many ways you can buy one fruit if all the mangoes are identical, all the apples are identical, & also all the Bananas are identical ?

Accepted Answer

As in this problem , buying any fruit is different case , as buying apple is independent from buying banana. so ADDITION rule will be used. C 1 9 + C 1 8 + C 1 6 = 23 will be answer.

Question 12

Find the sum of the all the numbers formed by the digits 2,4,6 and 8 without repetition. Number may be of any of the form like 2,24,684,4862 ?

Accepted Answer

Sum of 4 digit numbers = (2+4+6+8) x P 3 3 x (1111) = 20 x 6 x 1111 = 133320 Sum of 3 digit numbers = (2+4+6+8) x P 2 3 x (111) = 20 x 6 x 111 = 13320 Sum of 2 digit numbers = (2+4+6+8) x P 1 3 x (11) = 20 x 3 x 11 = 660 Sum of 1 digit numbers = (2+4+6+8) x P 0 3 x (1) = 20 x 1 x 1 = 20 Adding All , Sum = 147320

Question 13

A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ? A) 24 - 1 B) 2425-1 C) (24-1)(23-1)25 D) None

Accepted Answer

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects. Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection. Hence, we can select 1 black ball from 4 black balls or 2 black balls from 4 black balls. or 3 black balls from 4 black balls. or 4 black balls from 4 black balls. Hence, number of ways in which we can select the black balls = 4C1 + 4C2 + 4C3 + 4C4 = 2 4 - 1 ........(A) Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection. Hence, we can select 1 red ball from 3 red balls or 2 red balls from 3 red balls or 3 red balls from 3 red balls Hence, number of ways in which we can select the red balls = 3C1 + 3C2 + 3C3 = 2 3 - 1 ........(B) Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there) or 1 blue ball from 5 blue balls or 2 blue balls from 5 blue balls or 3 blue balls from 5 blue balls or 4 blue balls from 5 blue balls or 5 blue balls from 5 blue balls. Hence, number of ways in which we can select the blue balls = 5C0 + 5C1 + 5C2 + … + 5C5 = 2 5 ..............(C) From (A), (B) and (C), required number of ways = 2 5 2 4 - 1 2 3 - 1

Question 14

A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ?

Accepted Answer

The total number of ways of forming the group of ten representatives is ²²C₁₀. The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way The required number of ways = ²²C₁₀ - 1

Question 15

The number of permutations of the letters of the word 'MESMERISE' is ?

Accepted Answer

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways. The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R. Number of arrangements = 9 ! ( 2 ! ) 2 × 3 !

Question 16

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

Accepted Answer

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways = 600 ways.

Question 17

In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?

Accepted Answer

Given word is GLACIOUS has 8 letters. => C is fixed in one of the 8 places Then, the remaining 7 letters can be arranged in 7! ways = 5040.

Question 18

How many 4-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

Here given the required digit number is 4 digit. It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available. x x x 5 The remaining places can be filled by remaining digits as 5 x 4 x 3 ways. Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

Question 19

From a pack of 52 cards, 3 cards are drawn together atrandom, What is the probability of both the cards are king?

Accepted Answer

n(S) = 52C3 = 132600/6 = 22100 n(E) = 4C3 = 24/6 = 4

Question 20

In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person betwe…

Accepted Answer

A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. Required number of permutations = 18 x (17!) x 2 = 2 x 18!

Permutation and Combination Questions