Question 1

In the below word how many words are there in which R and W are at the end positions? RAINBOW

Accepted Answer

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways. Thus the total number of permutations = 2 x 5! = 2 x 120 = 240

Question 2

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

Accepted Answer

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210×6= 1260

Question 3

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

Accepted Answer

In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters =7!/2!= 2520. Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways. Required number of ways = (2520 x 20) = 50400.

Question 4

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Accepted Answer

Required number of ways = 7 C 5 × 3 C 2 = 63

Question 5

How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated ?

Accepted Answer

Three digit number will have unit’s, ten’s and hundred’s place. Out of 5 given digits any one can take the unit’s place. This can be done in 5 ways. ... (i) After filling the unit’s place, any of the four remaining digits can take the ten’s place. This can be done in 4 ways. ... (ii) After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits. This can be done in 3 ways. ... (iii) So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

Question 6

Find the number of ways to arrange 4 people in groups of 3 at a time where order matters?

Accepted Answer

P(4,3)= P 3 4 = 24

Question 7

Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?

Accepted Answer

6P4 = 6! / (6-4)! = 360

Question 8

Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter.?

Accepted Answer

C 5 20 = 15504

Question 9

How many ways are there to select a subcommittee of 7 members from among a committee of 17?

Accepted Answer

Since it does not matter what order the committee members are chosen in, the combination formula is used. Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation. C(17,7)= 19,448

Question 10

Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

Accepted Answer

When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used. C(52,5) = 2,598,960

Question 11

A school has scheduled three volleyball games, two soccer games, and four basketball games. You have a ticket allowing you to attend three of the games. In how many ways can you go to two basketball games and one of the other events?

Accepted Answer

Since order does not matter it is a combination. The word AND means multiply. Given 4 basketball, 3 volleyball, 2 soccer. We want 2 basketball games and 1 other event. There are 5 choices left. C(n,r) C(How many do you have, How many do you want) C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) C(4,2) x C(5,1) = (6)(5) = 30 Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.

Question 12

Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and ending with a 8?

Accepted Answer

There are 5 slots. __ __ __ __ __ The first slot must be a four. There are 4 ways to put a four in the first slot. There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc. (4)(4)(4)(4)(4) = 1024 Therefore there are 1024 different ways to produce the desired hand of cards.

Question 13

A certain marathon had 50 people running for first prize, second, and third prize.How many ways are there to correctly guess the first, second, and third place winners?

Accepted Answer

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

Question 14

How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000?

Accepted Answer

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9. 1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7) You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once. In each of these numbers, 7 is written once. Therefore, 243 times. 2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77 In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7). There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers. In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times. 3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it. Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300

Question 15

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 Or 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 in 8 C 3 ways. Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8 C 4 ways. Therefore, the total number of ways in which 8 students can travel is 8 C 3 + 8 C 4 = 56 + 70 = 126.

Question 16

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 17

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 18

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 19

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 20

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

Permutation and Combination Questions