PERMUTATIONS AND COMBINATIONS

INTRODUCTION TO PERMUTATIONS AND COMBINATIONS

1. Permutations : The ways in which a number of given objects can be arranged by taking all of them or a specified number of objects out of them are called PERMUTATIONS. Thus the number of permutations of three objects, viz. a, b, and c, taking all of them at a time is 6 i.e., abc, acb, bcd, bac, cab and cba. The number of ways in which 2 objects can be taken and arranged out of 3 objects a, b and c is 6, viz. ab, ba, bc, cb, ac and ca. The number of permutations of r things our of n things is denoted by nPr. Formula for Permutation : 1. nPr = n(n-1) (n-2) (1) n(n-1) (n-2) .1 is denoted by n! . 2. 0! =1 3. nPr =n! / (n-r)! 4. The number of ways in which n objects can be arranged in a circle is (n-1)! 2. Combinations The ways in which a specified number of objects can be taken out of a given number of objects (without regard to their arrangements) are called Combinations.The symbol nCr denotes the number of combinations or r things out of n things. Formula for nCr : nCr = n! / r! * (n-r)! Example 10C4 = (10 9 8 7)/ (1 2 3 4) Other important formulas: 1. nCr = nCn-r 2. nCr + nCr-1 = (n+1)Cr 3. nC0 + nC1 + nC2 + nC3 + .. nCn = 2n

PROBLEMS ON PERMUTATION AND COMBINATION

1) How many multiples of 5 are there from 10 to 95 ? Ans: 18 Explanation: As you know, multiples of 5 are integers having 0 or 5 in the digit to the extreme right. The first digit from the right can be chosen in2 ways. Thesecond digit can be any one of 1,2,3,4,5,6,7,8,9 i.e. There are 9 choices for the second digit. Thus, there are 2 x 9 = 18 multiples of 5 from 10 to 95. 2) In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A,B,C,D,E and F. How many different bus routes are possible? Ans: 594 Explanation: The number can be any one of the natural numbers from 1 to 99. There are 99 choices for the number. The letter can be chosen in 6 ways. Number of possible bus routes are 99 x 6 = 594 .

3) Evaluate 30!/28! Ans: 870 Explanation: 30!/28! = 30 * 29 * (28!) / (28!) = 30 * 29 =870 4) Find the value of 60P3 Ans: 205320 Explanation: 60P3 = 60! / (60 - 3)! = 60! / 57! = (60 * 59 *58 * (57!) )/ 57! = 60 * 59 *58= 205320 5) Find the value of 100C98 Ans: 4950 Explanation: 100C98 = 100C(100-98) = 100 * 99 / 2 *1 = 4950

6) In how many ways can a cricket eleven be chosen out of a batch of 15 players Ans: 1365 Explanation: Required number of ways = 15C 11 = 15C (15-11) = 15 C 4 15C 4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1= 1365 7) How many 4-letter word with or without meaning can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not allowed ? Ans: 5040 Explanation: 'LOGARITHMS' contains 10 different letters Required number of words = Number of arrangements of 100 letters taking 4 at a time = 10P4 = 10 * 9 * 8 * 7 = 5040

8) In how many ways can the letter of word 'LEADER' be arranged? Ans: 360 Explanation: The word 'LEADER' contains 6 letters namely 1L,2E,1A,1D and 1R Required number of ways = 6! / (1!)(2!)(1!)(1!)(1!) = 6 * 5 * 4 * 3 * 2 *1 / 2 * 1 =360 9) In how many different ways can the letter of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions

Ans: 36 Explanation: These are 6 letters in the given word , out of which there are 3 vowels and 3 consonants Let us mark these positions as under (1)(2) (3) (4)(5)(6) now 3 vowels can be placed at any of the three places out of 4 marked 1,3,5 Number of ways of arranging the vowels = 3P3 = 3! =6 Also,the 3 consonants can be arranged at the remaining 3 positions Number of arrangements = 3P3 = 6 Total number of ways = (6 * 6) =36 10) How many 3 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? Ans: 20 Explanation: Since each desired number is divisible by 5, so we much have 5 at the unit place. The hundreds place can now be filled by any of the remaining 4 digits .so, there 4 ways offilling it. Required number of numbers = (1 * 5 * 4) = 20 11) In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a self so that two books on Hindi may not be together? Ans: 1540 Explanation: In order that two books on Hindi are never together, we must place all these books as under: XEXEX..........XEX Where E denotes the position of an English and X that of a Hindi book. Since there are 21 books on English,the number of places marked X are therefore 22. Now, 19 places out of 22 can be chosen in 22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1 Hence the required number of ways = 1540 12) Out of 7 constants and 4 vowels how many words of 3 consonants and 2 vowels can be formed? Ans: 25200 Explanation: Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = 7C3 * 4C2 = 210 Number of groups each having 3 consonants and 2 vowels = 210 Each group contains 5 letters Number of ways of arranging 5 letters among themselves = 5! = (5 * 4 * 3 * 2 * 1) = 210 Required number of words = (210 * 210) = 25200

13) How many arrangements can be made out of the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together Ans: 120960 Explanation: In the word ' MATHEMATICS' we treat vowels AEAI as one letter thus we have MTHMTCS(AEAI) nowwe have to arrange 8 letters out of which M occurs twice ,T occurs twice & the rest are different Number of ways of arranging these letters = 8! / (2!)(2!) = 10080 now AEAI has 4 letters in which A occurs 2 times and the rest are different Number of ways of arranging these letters = 4! / 2! = 12 Required number of words = (10080* 12) = 120960 14) Find the number of ways to arrange 4 people in groups of 3 at a time where order matters. Ans: 24 Explanation: There are 24 ways to arrange 4 items taken 3 at a time when order matters. 15) Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter. Ans: 15,504 Explanation:

There are 15,504 ways to arrange 20 objects taken 5 at a time when order does not matter. 16) Is the State Lotteries a combination or a permutation? Ans: The answer has to do with if order matters. If order matters, then it is a permutation. If order does not matter then it is a combination. Do you think the numbers on a ticket have to be in the same order as the order in which they became the winning numbers? In other words, lets say the winning numbers rolled out of the machine in the order of: 1,2,3,4,5,6. Do the numbers on your ticket have to be in this same order to win? Or will any order such as 2,3,1,5,6,4 also be a winning ticket? The answer is, any order of the winning numbers will produce the winning ticket. Thus the lotteries are combinations. 17) Determine the total number of five-card hands that can be drawn from a deck of 52 cards. Ans: 2,598,960 Explanation: 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

18) 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? Ans: 30 Explanation: 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)*C(have 5 choices left, want 1) C(4,2)*C(5,1) (6)(5) = 30 19) How many ways are there to deal a five-card hand consisting of three eight's and two sevens. Ans: 24 Explanation: If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed. The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards. We have 4 eights and 4 sevens. We want 3 eights and 2 sevens. C(have 4 eights, want 3 eights)*C(have 4 sevens, want 2 sevens) C(4,3)*C(4,2) = 24 20) How many different 5-card hands include 4 of a kind and one other card? Ans: 624 Explanation: We have 13 different ways to choose 4 of a kind: 2's, 3's, 4's, Queens, Kings and Aces. Once a set of 4 of a kind has been removed from the deck, 48 cards are left. Remember OR means add. The possible situations that will satisfy the above requirement are:4 Aces and one other card C(4,4)*C(48,1) = 48 or 4 Kings and one other card C(4,4)*C(48,1) = 48 or 4 Queens and one other card C(4,4)*C(48,1) = 48 . . . or 4 twos and one other card C(4,4)*C(48,1) = 48 Total of 624 ways This problem could also have been calculated as follows. 13*C(4,4)*C(48,1) = 624