Question 1

In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

Accepted Answer

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind. Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

Question 2

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

Accepted Answer

The first letter is E and the last one is R. Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters. The second and third positions can either have two different letters or have both the letters to be the same. Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways. Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways. Total number of possibilities = 56 + 3 = 59

Question 3

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

Accepted Answer

A triangle is formed by joining any three non-collinear points in pairs. There are 7 non-collinear points The number of triangles formed = 7 C 3 = 35

Question 4

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

Accepted Answer

Possibilities Bowlers Batsmen Number of ways 6 9 1 4 7 ( 6 C 4 * 9 C 7 ) 2 5 6 6 C 5 * 9 C 6 3 6 5 6 C 6 * 9 C 5 6 C 4 * 9 C 7 = 15 x 36 = 540 6 C 5 * 9 C 6 = 6 x 84 = 504 6 C 6 * 9 C 5 = 1 x 126 = 126 Total = 1170

Question 5

Find the value of 'n' for which the nth term of two AP'S: 15,12,9.... and -15,-13,-11...... are equal?

Accepted Answer

Given are the two AP'S: 15,12,9.... in which a=15, d=-3.............(1) -15,-13,-11..... in which a'=-15 ,d'=2.....(2) now using the nth term's formula,we get a+(n-1)d = a'+(n-1)d' substituting the value obtained in eq. 1 and 2, 15+(n-1) x (-3) = -15+(n-1) x 2 => 15 - 3n + 3 = -15 + 2n - 2 => 12 - 3n = -17 + 2n => 12+17 = 2n+3n => 29=5n => n= 29/5

Question 6

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

Accepted Answer

Let 4 girls be one unit and now there are 6 units in all. They can be arranged in 6! ways. In each of these arrangements 4 girls can be arranged in 4! ways. Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

Question 7

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

Accepted Answer

The 7 letters word 'POVERTY' be arranged in 7 P 7 ways = 7! = 5040 ways.

Question 8

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

Accepted Answer

Total number of letters in the word ABYSMAL are 7 Number of ways these 7 letters can be arranged are 7! ways But the letter is repeated and this can be arranged in 2! ways Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

Question 9

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

Accepted Answer

Let the beds be numbered 1 to 7. Case 1 : Suppose Anju is allotted bed number 1. Then, Parvin cannot be allotted bed number 2. So Parvin can be allotted a bed in 5 ways. After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways. So, in this case the beds can be allotted in 5´5!ways = 600 ways. Case 2 : Anju is allotted bed number 7. Then, Parvin cannot be allotted bed number 6 As in Case 1, the beds can be allotted in 600 ways. Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin. Therefore, Parvin can be allotted a bed in 4 ways in all these cases. After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways. Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

Question 10

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

Accepted Answer

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating. The first 24 of these words will start with A. Then the 25th word will start will CA _ _ _. The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA. The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM. Therefore, the rank of CHASM will be 24+6+2= 32

Question 11

How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ?

Accepted Answer

Let last digit is 2 when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways. so for last digit = 2, total numbers=240 Similarly for 4 and 6 When last digit = 4, total no. of ways =240 and last digit = 6, total no. of ways =240 so total of 720 even numbers are possible.

Question 12

The number of ways in which 8 distinct toys can be distributed among 5 children?

Accepted Answer

As the toys are distinct and not identical, For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are 5 8 ways to distribute the toys. Hence, it is 5 8 and not 8 5 .

Question 13

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?

Accepted Answer

Given word is THERAPY. Number of letters in the given word = 7 Number of vowels in the given word = 2 = A & E Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is 6! x 2! = 720 x 2 = 1440.

Question 14

Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

Accepted Answer

Here the order of choosing the elements doesn’t matter and this is a problem in combinations. We have to find the number of ways of choosing 4 elements of this set which has 11 elements. This can be done in 11 C 4 ways = 330 ways

Question 15

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are to occupy only odd places.

Accepted Answer

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants. There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4 P 3 ways and 4 constants can be arranged in 4 P 4 ways. Number of words = 4 P 3 x 4 P 4 = 24 x 24 = 576

Question 16

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?

Accepted Answer

From 5 consonants, 3 consonants can be selected in 5 C 3 ways. From 4 vowels, 2 vowels can be selected in 4 C 2 ways. Now with every selection, number of ways of arranging 5 letters is 5 P 5 ways. Total number of words = 5 C 3 * 4 C 2 * 5 P 5 = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

Question 17

There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?

Accepted Answer

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room, Then, 7C1 x 6C2 x 4C4 = 7 x 15 x 1 = 105

Question 18

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

Accepted Answer

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O. There are 6 letters in thegiven word. First arrange 3 vowels. This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O) Remaining 3 letters can be placed in 3 places = 3! ways Total number of possible ways of arranging letters of OLIVER = 3! x C 3 6 ways = 6x5x4 = 120 ways.

Question 19

Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?

Accepted Answer

Number of words with 5 letters from given 9 alphabets formed = 9 5 Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 9 P 5 Number of words can be formed which have at least one letter repeated = 9 5 - 9 P 5 = 59049 - 15120 = 43929

Permutation and Combination Questions