PERMUTATION AND COMBINATION

Q1. Find the number of even natural numbers, which have three digits?

Ans1. The total number of digits are 10 (0 to 9), The three digit number has three places to fill. The first place can be filled by 9 (excluding zero), the second by all 10 and to make it even the third has can only be filled by 5 (0, 2, 4, 6, 8).

9 10 5

Total three digit even numbers = 9x10x5 = 450

Students should not get confused here with the concept of repeat and non-repeat usage of digits, since the question asks for the three digits numbers present in the natural numbers. Now in natural numbers, numbers are formed from all digits with repetitions, so there should be no confusion.

Q2. In how many different ways six questions of true false type can be answered?

Ans2. Each question can be answered in two ways – true or false, which is 2 ways, as total number of questions is six, so all six can be answered in 2x2x2x2x2x2 = 64 ways

Q3. In how many different ways six questions of true false type can be answered incorrectly?

Ans3. Here incorrectly means, the options which are other than where all the answers are correct. The option where all the answers are correct is only 1. Since total number of ways of answering is 64(from last example). So total no. of ways are 64 - 1 = 63 ways

Q4. Find the number of permutations of the letters of the word custom such that no repetitions are there. How many words beginning with M? How many digits begin with M and end with S?

Ans4. The word “custom” has all distinct alphabets, no repeats. The permutation of 6 objects to be placed in six places is 6! = 720 ways

If the word has to begin with M, then first place is locked with M, so there are five places left to be filled with five alphabets = 5! = 120 ways

If the word has to begin with M, and end with S, then these two positions are locked, so there are four places to be filled with four alphabets = 4! = 24 ways

Q5. Find the number of ways in which the letters of the word “epidemic” can be arranged?

Ans5. The word “epidemic” has total 8 alphabets with “i” repeating two times and “e” repeating two times, so by the formula n! / (p! q! r!..), since there are two repeats, the number of ways is

8!/(2!2!) = 10080

Q6. Find the number of ways in which the letters of the word India can be arranged?

Ans6. The word “India” has total 5 alphabets with “i” repeating two times, so by the formula n! / (p! q! r!..), since there is one repeat, the number of ways is

5!/2! = 60

Q7. In how many ways can the six letters A, B, C, D, E, F can be arranged such that B and C always come together?

Ans7. Since B and C have to together, therefore it becomes set of 5 things

5 places 5 things is given by 5P₅ ways = 120 ways

Now B and C can arrange themselves in two ways BC and CB

Therefore total ways = 120 x 2 = 240 ways

Q8. If Nirula’s offers 31 flavours of ice cream cones in sizes small, medium and large, how many different selections of cones are possible? if 4 toppings are available to put on a cone, and a cone can be bought without a topping, how many different selections of ice cream cones are available?

Ans 8 Since there are 31 flavours and three sizes, the total number of selections will be = 31 x 3 = 93.

If there are four toppings, they can be taken in 2⁴, since there are 93 ways of taking the ice-cream itself total number of ways = 93 x 2⁴ = 1488

Q9. How many 8 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels?

Ans9. Out of eight spaces 3 are to be reserved for vowels, they can be selected in 8C₃ ways. Now there are 5 vowels and each can take any of the three positions = 5³ and there are 21 consonants which can take rest of the 5 positions = 25⁵

Total number of ways = 8C₃ x 5³ x 25⁵ = 28588707000 ways

Q10. In how many ways can 5 boys and 3 girls be made to stand in a row such that no two girls are together?

Ans10. Since no two girls can stand together they take places between the boys, before after and between, there are six positions around 5 boys, where three girls have to fit, therefore it is 6P₃. Now boys are left to take 5 positions (for 5 places) which is 5!

Total ways = 6P₃ + 5! = 14400 (why add, as both the things are independent)

(permutationcombination theory Fig 3)

Q11. In how many ways can a committee of 3 men and 3 ladies be appointed from 6 men and 4 ladies?

Ans11. 6 Men, three to be selected, which is 6C₃

4 Ladies, two to be selected, which is 4C₂

Total ways of selection = 6C₃ x 4C₂ = 120

Q12. From 6 men and 4 ladies five people have to be chosen with at least one lady?

Ans12. 6 Men + 4 Ladies, 10 people, five to be selected, which is 10C₅

The scenario in which no women is selected, 6 men, 5 to be selected 6C₅

Here 10C₅ - 6C₅ is where at least one lady is selected = 246 ways

Q13. In a conference of 9 schools, how many intra conference football games are played during the season if the teams all play each other exactly once?

Ans13. There are total of nine teams and two teams required to play a match, so the total number of ways is 9C₂ = 36 games

Q14. How many different signals can be made by using at least three distinct flags if there are five different flags from which to select?

Ans14. At least three distinct flags are to be chosen from give 5 flags, which is:

5P₃ + 5P₄ + 5P₅ = 5!/2! + 5!/1! + 5!/0! = 300 signals

Q15. In how many ways 11 cricketers can be chosen from 6 bowlers , 4 wicket keepers and 11 batsmen to give a majority of batsmen if at least 4 bowlers are to be included and there is one wicket keeper?

Ans 15. One wicketkeeper has to be choose out of 4 which is 4C₁ = 4

Now since 11 players have to be selected, post wicketkeeper selection only 10 are left, out of which at least 4 bowlers have to be selected and batsman should have majority, so they will be 6, and that’s the only way.

Ways of selecting four bowlers out of 6 = 6C₄

Ways of selecting Six batsmen out of 11 = 11C₆

Total ways = 11C₆ x 6C₄ x 4 = 27720

Q16. If 6 balls of different colors - black, white, yellow, green, blue, violet are to be arranged in a row that the black and white balls may never come together?

Ans16. There are 6 balls, in case black and while ball come together, then the scenario will be of 5 objects and 5 places = 5P₅ x 2 (since black and white can change positions) = 240

Total arrangements = 6P₆ = 6! = 720

Arrangements where black and white ball are not together = total arrangements – the case black and while ball come together = 720 – 240 = 480

Q17. A Teacher with 8 students takes three at a time to the computer room, as often as he can without taking the same three students together more than once. How often will he go, and how often will each student go?

Ans17. The visits of the teacher = number of groups, as he visits with each group

Which is three students out of eight = 8C₃ = 56

The student cannot be paired with the other two of his group, therefore he will go as many times as the other two in his group can be taken out of the remaining 7 students other than him = 7C2 = 21

Think about the second point, you will get it.

Q18. A team consists of 8 men, 3 of whom can only work in one city and 2 only in the other. Find the number of ways in which the team can be formed. There have to be equal number of men in both cities. Also find the number of ways in which the team can be formed and internally rearranged.

Ans18. If 3 are fixed in one city and 2 are fixed in other, out of eight five positions are closed. Since teams have to be of 4 people each, one position is open in one city and two positions are open in two cities. So there are three positions and three people, since there is only one position in one city, only one can fitted there, and all three people can be fitted there, those are the only three ways. Therefore there are 3 ways of making the teams.

The two teams have four members each and both can be arranged in 4! Ways = 24

Total ways = 24 x 24 x 3 = 1728.

Q19. A Bus goes from Delhi to Udaipur(last stop) stops at 8 intermediate terminals. 5 persons board the bus during the journey with 5 different tickets. How many different sets of tickets may they have had?

Ans19. There are 8 intermediate terminals, if a person boards from first terminal after Delhi, he can buy 8 type of tickets up to Udaipur. Similarly the next person boarding from next terminal can buy 7 types of tickets and so on.

So type of tickets = 8+7+6+5+4+3+2+1 = 36

Now the five people who boarder the bus may have any of the 36 type of tickets = 36C₅ = 376992 sets

Q20. Find the number of diagonals in a decagon.

Ans20. The number of sides in a decagon is 10; the number of ways two sides can meet are 10C₂ = 45, there are 10 outer sides, therefore 45 -10 = 35

Q21. Find the number of triangles in an octagon.

Ans21. The number of sides in an octagon is 8; the number of ways three sides can meet to form a triangle are 8C₃ = 56, there are 56 triangles

Q22. There are 20 points in which 6 points which are collinear. How many straight lines can be formed by joining them?

Ans22. Total lines (meeting of two points) for 20 points = 20C₂

Collinear points = 6, therefore only one line for 6C₂

Therefore total lines = 20C₂ - 6C₂ + 1 = 176 lines

Q23. If you have 100 people at a party, and everyone shakes hands with everyone else, how many handshakes take place?

Ans23. There are two people required for a handshake therefore 100C₂ = 4950

Q24. Four figures are to be inserted into a six-page essay, in a given order. One page may contain at most two figures. How many different ways are there to assign page numbers to the figures under these restrictions?

Ans24. In the case that one page may contain 1 figure, the number of ways are 6C4. In the case where 1 page only can contain 2 figures, the ways are 6C3 Consider 2 pages can contain 2 figures then 6C2. The total number of ways are 6C4 + 6C3 + 6C2 = 50 ways.

Q25. There are n Railway stations. Ticket facility is 10 available between every two stations. Recently, m new stations are built so that 42 new tickets are to be printed. How many stations were there? How many are newly Constructed.

Ans25. For the m new stations you will need to print 42 new tickets = 21 new tickets one way. Now, 21 = trains between 1 each of n stations and 1 of m stations = m × n + trains between each of the m new stations = mC2

21 = m*n + mC2 = m(2n + m – 1)/2

3*7 = m(2n + m–1)/2

Taking m = 3

7 = (2n + 3–1)/2

14 = 2n + 2

n = 6

If we take m = 7, (2n + 6) = 6  n = 0

Therefore, there were no stations initially and 7 new stations were added or 6 originally, 3 new were added.

Q26. How many ways can you buy a dozen donuts from an unlimited supply of 5 types of donuts?

Ans26. The way here is to think of how many ways can you line up 12 x’s and 4/’s. Why? There is a one to one correspondence between such lineups and possible purchases — xx/xxx//xxxxxx/x corresponds to 2 of type 1, 3 of type 2, 0 of type 3, 6 of type 4 and 1 of type 5 etc. Thus we need to count the number of such lineups. Equivalently, how many ‘words’ can we make from 4/’s and 12 x’s? This is fairly easy as we have 16 spots to fill and 4 of them have to be chosen to be occupied by a/. Thus there are 16C4*12C12 = [16]C[4] = 1820 ways to buy the donuts.

What if the question says that you must purchase at least one of each type? Then the answer is only 11C4*7C7 = 11C4 ways.

Q27. A company president is deciding how to fill three vice-presidencies in the company: VP-Marketing, VP-Finance, and VP-Production. Twelve executives are eligible and qualified for promotion, and each could fill any of the three positions. In how many ways can the positions be filled?

Ans27. The decision-maker first selects three people from among the twelve, not yet thinking about their job assignments (order not important). This can be done C(12,3) = 220 ways. Then the decision-maker assigns the three chosen people to the three jobs (order important). This can be done P(3,3) = 6 ways. So the total number of ways is 220 x 6 = 1,320.

Q28. The Lottery Commission is considering a new game in which five balls would be withdrawn from a box containing 10 balls, numbered 0 to 9. The five balls would come out of the box at nearly the same time, as they do in the current Lotto game, in which six balls come out of a box into a tube at nearly the same time. In this new game, however, the winning ticket must have the five lucky numbers in the same order as they came out of the box. What is the chance of winning with a single five-number ticket?

Ans28. The order is important, duplicates are not possible. Therefore 10P5 = 30,240.

Formula-free sequential method: the first number has 10 possibilities, the second number has 9, the third number has 8, the fourth number has 7, and the fifth number has 6. 10 × 9 × 8 × 7 × 6 = 30, 240. So the probability of winning is 1/30,240 = 0.000033069.

Q29. A new flag is to be designed with six vertical stripes using some or all of the colors yellow, green, blue and red. Then, find the number of ways this can be done such that no two adjacent stripes have the same color. (CAT 2003).

Ans29. The first stripe can be chosen as 4C1 = 4 ways, as one out of four colors is to be taken, rest all are 3C1 = 3 ways as the adjacent color cannot be repeated, which leaves one color to be chosen from three colors. Therefore total ways = 4 × 3 × 3 × 3 × 3 × 3 = 972 ways

Q30. What is the maximum number of points of intersections of:

(a) Five circles

(b) Five straight lines

(c) Three circles and three straight lines

Ans30. (a) For circles permutation will be used, as arrangement is important 5P2 = 20

(b) For lines arrangement is irrelevant, therefore use of combination 5C2 = 10

(d) Three circles = 3P2 = 6, three line = 3C2 = 3, Now three lines can maximum cut three circle in 18 times(as one line can cut three circles 6 times), so total intersections = 18 + 6 + 3 = 27

Q31. In a Ranji Cup final, team A and team B play until one team wins 4 games. The sequence of game winners is designated by letter; for example, ABBBB means team A won the first game and team B won the next four games. How many different Ranji Cup finals are possible?

Ans31. The number of ways with B winning is shown in the table.

Combinations Number of Sequences BBB|B 1

ABBB|B 4!/[1!3!] = 4

AABBB|B 5!/[2!3!] = 10

AAABBB|B 6!/[3!3!] = 20

Total = 35

we get a further 35 possible sequences with A winning, so the total number of sequences is 35 + 35 = 70 There are 70 possible sequences for the Ranji Cup finals.

Q32. Ravi’s family consists of a grandfather, ‘X’ sons and daughters and ‘Y’ grand children. They are to be seated in a row for dinner. The grandchildren wish to occupy the Y seats at each end and the grandfather refuses to have a grandchild on either side of him., In how many way’s can the Ravi’s family be made to sit?

Ans32. Total Seats = X + Y + 1

Grandchildren (Y) sit together on Y seats in Y! ways

Since grandfather does not want to sit with any grandchildren, he cannot take any of the Y seats and 2 seats on either side of Y seats. Which is X + 1 – 2 = X – 1 seats left for him to sit, so one person to be accommodated in X – 1 seats is done in X – 1 ways. The rest X people can sit on X seats in X!

So total ways = X! Y! (X-1)

Q33. There are 720 (or 6!) permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged from smallest to largest numerical value, beginning with 123456 and ending with 654321, then what number falls on the 124th position?

Ans33. Among the 6! Permutations, there are 5! = 120 beginning with 1. They would be numbered from 1 to 120. The next 120, including the 124th, would begin with 2. Among the 5! Permutations beginning with 2, there are 4! = 24, including the 124th, which have second digit 1. Among the 4! Permutations beginning with 21, there are 3! = 6, including the 124th, which have third digit 3. Among the 3! Permutations beginning with 213, there are 2! = 2 with fourth digit 4 (numbers 121 and 122), 2 with fourth digit 5 (numbers 123 and 124), and 2 with fourth digit 6 (numbers 125 and 126). The 124th has fourth digit 5. Finally, among the 2! Permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5th digit 6 (number 124). The 124th is the latter one. Thus the 124th number is 213564.

Q34. Find the number of ways’ in which 9 identical balls can be placed in three identical drums.(assuming each drum can take all nine balls)

Ans34. The balls can go in any drum, and their being in one drum or another does not make a difference since both balls and drums are identical. This is the important point to remember. Now, one situation is that all nine balls in one drum, again remember it does not matter which drum they are in, that’s one way, in this way the other two drums are empty. Now, if one drum has no ball, then other drums could have four combinations to make nine (as total is 9) out of two digits (8, 1), (7, 2), (6, 3), (5, 4), now these are four ways. Now, if one drum has one ball, then other drums could have four combinations to make eight (as total is 9) out of two digits(7, 1), (6, 2), (5, 3), (4, 4), now these are four ways. Now, if one drum has two balls, then other drums could again have three combinations to make seven (as total is 9) out of two digits (6, 1), (5, 2), (4, 3), out of which (6, 1) is a repetition of one of the previous cases, so in all two ways. Finally the only way in which all drums have three balls. So total cases = 1 + 4 + 4 + 2 + 1 = 12 ways

Q35. In how many ways a cricketer can make a century with fours and sixes only?

Ans35. Taking the extreme case, with all fours(25) he can make a century, with only sixes he cannot do, as 100 is not a multiple of 6, so with 16 sixes and one four he can do it. Now from here keep reducing six and see if the number left is a multiple of four that’s a case. The next case is 14 sixes and 4 fours, the next is 12 sixes and 7 fours, and so on, you will find that there are 9 ways in all.

Q36. At a party, the superintendent of an apartment building tells you that his building has seven elevators. Each elevator stops on at most 3 floors. He also tells you that if you take the right elevator, you can get to any one floor from any other floor without changing elevators. What is the greatest number of floors that the building could have?

Ans36. Since the elevator stops at three floors, we can make a combination of the same:

1 2 3

1 4 5

1 6 7

2 4 6

2 5 7

3 4 7

3 5 6

Therefore for one elevator there could be 7 floors, so for seven elevators there could be maximum of 21 floors (7 × 3)