Question 1

Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first c…

Accepted Answer

Since two of the eight first class petty officers are to fill two different offices, we write 8 P 2 = 56 Then, two of the six second class petty officers are to fill two different offices; thus, we write 6 P 2 =30 The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

Question 2

There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?

Accepted Answer

fix one person and the brothers B1 P B2 = 2 ways to do so. other 17 people= 17! Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!

Question 3

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

Accepted Answer

Total number of alphabets = 10 so ways to arrange them = 10! Then there will be duplicates because 1st S is no different than 2nd S. we have 4 Is 3 S and 2 Ps Hence number of arrangements = 10!/4! x 3! x 2! = 12600

Question 4

In how many ways, can zero or more letters be selected form the letters AAAAA?

Accepted Answer

Selecting zero'A's= 1 Selecting one 'A's = 1 Selecting two 'A's = 1 Selecting three 'A's = 1 Selecting four 'A's = 1 Selecting five 'A's = 1 => Required number ofways =6

Question 5

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

Accepted Answer

Two horses A and B, in a race of 6 horses... A has to finish before B if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4! if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4! if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4! if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways.. A cannot finish 6th, since he has to be ahead of B Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

Question 6

A committee of 4 people is to be formed from a group of 9 people.How many possible committees can be formed?

Accepted Answer

This question is a combination since order is not important. Answer = 7 C 3 = 126

Question 7

A pizza can have 3 toppings out of possible 7 toppings.How many different pizza's can be made?

Accepted Answer

There are 7 toppings in total,and by selecting 3,we will make different types of pizza. This question is a combination since having a different order of toppings will not make a different pizza. So, 7 C 3 =35

Question 8

How many ways can you select 17 songs for mix CD out of possible 38 songs?

Accepted Answer

since it is a combination = 38 C 17 = 2878 x 10^10

Question 9

A school committee of 5 is to be formed from 12 students.How many committees can be formed if John must be on the committee?

Accepted Answer

If John must be on the committee,we have 11 students remaining,out of which we choose 4. So, 11 C 4

Question 10

From a deck of 52 cards, a 7 card hand is dealt.How many distinct hands are there if the hand must contain 2 spades and 3 diamonds ?

Accepted Answer

There are 13 spades,we must include 2: 13 C 2 There are 13 diamonds,we must include 3: 13 C 3 Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26 C 2 Therefore, 13 C 2 * 13 C 3 * 26 C 2 = 7250100

Question 11

From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?

Accepted Answer

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3. So, 50 C 3

Question 12

In how many ways can you choose one or more of 12 different candies?

Accepted Answer

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 2 12 . Since the case of no candy being selected is not an option, we have to subtract 1 from our answer. Therefore,there are 2 12 - 1 = 4095 ways of selecting one or more candies

Question 13

How many 3 letters words can be formed using the letters of the words hexagon?

Accepted Answer

Since the word hexagon contains 7 different letters,the number of permutations is 7 P 3 = 7 x 6 x 5 =210

Question 14

In how many different ways can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

Accepted Answer

Number of permutations = 4 P 3 = 24

Question 15

When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patients at a time,find the number of ways, he can schedule his patients if 3 leave in disgust before Dr. Ram gets around to seeing them.

Accepted Answer

There are 12 - 3 = 9 patients.They can be seen in 12 P 9 = 79833600 ways

Question 16

In how many ways can 4 sit at a round table for a group discussions?

Accepted Answer

We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6

Question 17

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

Accepted Answer

The toys are different; The boxes are identical If none of the boxes is to remain empty, then we can pack the toys in one of the following ways a. 2,2,1 b. 3,1,1 Case a. Number of ways of achieving the first option 2−2–1 Two toys out of the 5 can be selected in 5 C 2 5 ways. Another 2 out of the remaining 3 can be selected in 3 C 2 ways and the last toy can be selected in 1 C 1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2. Therefore, total number of ways of achieving the 2−2–1 option is: 5 C 2 * 3 C 2 2 = 10 * 3 2 = 15 ways. Case b. Number of ways of achieving the second option 3−1–1 Three toys out of the 5 can be selected in 5 C 3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3−1–1 option is 5 C 3 =10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes =number of ways of achieving Case a + number of ways of achieving Case b =15 + 10 = 25 ways

Question 18

In how many different ways can 6 different balls be distributed to 4 different boxes, when each box can hold any number of ball?

Accepted Answer

Every ball can be distributed in 4 ways. Hence the required number of ways = 4 x 4 x 4 x 4 x 4 x 4 = 4096

Question 19

What is the total no of ways of selecting atleast one item from each of the two sets containing 6 different items each?

Accepted Answer

We can select atleast one item from 6 different items = 2 6 - 1 Similarly we can select atleast one item from other set of 6 different items in 2 6 - 1 ways. Required number of ways = ( 2 6 - 1 ) 2 6 - 1 = 2 6 - 1 2 = 3969

Question 20

16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

Accepted Answer

(16 – 2)! × 2 = 14! × 2

Permutation and Combination Questions