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In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions. PermutationsDefinition
For example: The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC. Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations. The number of ways in which n things can be arranged, taken all at a time, nPn = n!, called ‘n factorial.’ Factorial FormulaFactorial of a number n is defined as the product of all the numbers from n to 1. For example, the factorial of 5, 5! = 5*4*3*2*1 = 120. Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways. Number of permutations of n things, taken r at a time, denoted by: For example: The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways. Important Permutation Formulas1! = 1 0! = 1 Let us take a look at some examples: Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’. Solution: ‘CHAIR’ contains 5 letters. Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120. Solution: The word ‘INDIA’ contains 5 letters and ‘I’ comes twice. When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter. Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60. Solution: The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice. Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080. Solution: The word ‘SUPER’ contains 5 letters. In order to find the number of permutations that can be formed where the two vowels U and E come together. In these cases, we group the letters that should come together and consider that group as one letter. So, the letters are S,P,R, (UE). Now the number of words are 4. Therefore, the number of ways in which 4 letters can be arranged is 4! In U and E, the number of ways in which U and E can be arranged is 2! Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways. Solution: The word ‘BUTTER’ contains 6 letters. The letters U and E should always come together. So the letters are B, T, T, R, (UE). Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice). Number of ways in which U and E can be arranged = 2! = 2 ways Therefore, total number of permutations possible = 60*2 = 120 ways. Problem 6: Find the number of permutations of the letters of the word ‘REMAINS’ such that the vowels always occur in odd places. Solution: The word ‘REMAINS’ has 7 letters. There are 4 consonants and 3 vowels in it. Writing in the following way makes it easier to solve these type of questions. (1) (2) (3) (4) (5) (6) (7) No. of ways 3 vowels can occur in 4 different places = 4P3 = 24 ways. After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4P4 = 4! = 24 ways. Therefore, total number of permutations possible = 24*24 = 576 ways. CombinationsThe different selections possible from a collection of items are called combinations. For example: The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA. It does not matter whether we select A after B or B after A. The order of selection is not important in combinations. To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by nCr is nCr = n! / [r! * (n-r)!] For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are (adsbygoogle = window.adsbygoogle || []).push({}); 3C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)Important Combination formulasnCn = 1 nC0 = 1 nC1 = n nCr = nC(n-r) The number of selections possible with A, B, C, taken all at a time is 3C3 = 1 (i.e. ABC) Solved examples of CombinationLet us take a look at some examples to understand how Combinations work: Solution: No. of ways 1 man can be selected from a group of 3 men = 3C1 = 3! / 1!*(3-1)! = 3 ways. No. of ways 3 women can be selected from a group of 4 women = 4C3 = 4! / (3!*1!) = 4 ways. Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls. Solution: Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be 3 B and 2 R 4 B and 1 R and 5 B and 0 R balls. Therefore, our solution expression looks like this. Solution: If a number is divisible by 10, its units place should contain a 0. After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits. Selecting one digit out of 5 digits can be done in 5C1 = 5 ways. After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C1 = 4 ways. After filling the hundreds place, the thousands place can be filled in 3C1 = 3 ways. Therefore, the total combinations possible = 5*4*3 = 60. Permutations and Combinations QuizTry these practice problems. Problem 1: Click here
Solve the following. i) 30P2 A. 870, 435 B. 435, 870 C. 870, 470 D. 435, 835 Answer 1: Click here
A Explanation: 30P2 = 30! / 28! = 30*29*28! / 28! = 30*29 = 870 30C2 = 30! / (2!*28!) = 435 Problem 2: Click here
How many different possible permutations can be made from the word ‘BULLET’ such that the vowels are never together? A. 360 B. 120 C. 480 D. 240 Answer 2: Click here
D. Explanation: The word ‘BULLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360 No. of permutations possible with vowels always together = 5! * 2! / 2! = 120 No. of permutations possible with vowels never together = 360-120 = 240 Problem 3: Click here
In how many ways can a selection of 3 men and 2 women can be made from a group of 5 men and 5 women ? A. 10 B. 20 C. 30 D. 100 Answer 3: Click here
D. Explanation: 5C3 * 5C2 = 100 =3360. Let su assume LLL as 1 letter. Then, LLL + PARAE has 6 letters, out of which there are 2 A’s and the rest are all distinct. Number of their arrangements =6! How many different arrangements of the word are there?Therefore, we can arrange the letters in the word ‘FACTOR’ in 720 ways. Thus, this is the required answer. How many distinct permutations can be made from the word parallel?No. of Permutations=3360. How many words can be formed by arranging the letters of the word Mumbai?So the total number of words using the letters of MUMBAI such that all the M’s come together are 120. So this is the required answer. How many ways are there to arrange the letters of the word jethalal?Find the number of permutation of n distinct things taken r together ,in which 3 particular things must occur together . How many words can be formed by arranging the letters of the word computer the vowels never come together?Similarly, there is only 1 way to arrange the vowels in the remaining 3 positions. Therefore, there are 56 ways to arrange COMPUTER, given the above constraint. Alternatively we can choose to position the vowels first. We will get C(8,3) = 56 ways to select a combination of positions on which the vowels must stay. How many words can be formed using all the letters of the word train?34 words can be made from the letters in the word train. This page is a list of all the words that can be made from the letters in train, or by rearranging the word train. How many words can be formed with the letters of the word college?26 words can be made from the letters in the word college. How many arrangements are there for all letters in the word sociological?Since there are 60 Xi, and for each of them there are 1260 distinct permutations of the letters in SOCIOLOGICAL, in which all the vowels are adjacent, totally in 60. 1260 =75600 of the arrangements of the letters in SOCIOLOGICAL, all the vowels are adjacent. How many arrangements are there of all letters in mathematics?The word MATHEMATICS consists of 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I, 1 C and 1 S. Therefore, a total of 4989600 words can be formed using all the letters of the word MATHEMATICS. Therefore, a total of 453600 words which begin with C can be formed using all the letters of the word MATHEMATICS. How many new words are possible from the letters of the word permutation?There are 12 letters in “Permutations” – 1 P and 1 S (we’ll work with the other letters in a minute). and so on, for a total of 7 placements. We can also reverse the order of P and S, thus doubling the number of placements to 14. which gives 10×9×8×7×6×5×4×3×2×1=10! How many different ways the letters of the word algebra can be arranged in a row if a the two A’s are together by the two A’s are not together?question_answer Answers(1) In a word ALGEBRA have 2 A’s and 5 different letters are there. 1) Two A’s will take 1 unit and 5 letters will take 5 units then total number units = 6. These can be arranged in 6! = 6x5x4x3x2x1 = 720 ways. How many combination you can create by rearranging the alphabets of the word Indians so that the vowels always come together?= 120 ways. The vowels (EAI) can be arranged among themselves in 3! How many words can be made by arranging any 3 letter of Sunday?Answer is option b)720 W … How many letters are there in MUMBAI?The word MUMBAI consists of 6 letters taht include two Ms. When we consider both the Ms as a single entity, we are left with 5 entities that can be arranged in 5! ways. How many 4 letter words with or without meaning can be formed?Therefore, the number of four-letter words that can be formed is 5040. How many different arrangement can be made from the letters of word system such that the word begin with E?Hence, the total number of ways are 180. But out of these 4 letters E comes 2 times. How many different arrangements can be formed by using all letters of the word analysis?There are 15,120 ways. How many arrangements can be made by taking four letters of the word Mississippi?Here as some letters are being repeated and hence we can not simply go for P(11,4). Total number of 4 letter words formed from the letters of the word MISSISSIPPI can be computed by summing up the result of all these 5 cases. Therefore total of 176 words can be formed from the letters of the word MISSISSIPPI. How do you make a train in Word?A word train is a series of words in which each successive word starts with the first letter of the previous word. These can be made in any manner according to one’s vocabulary. How many words can be formed by using all letters of the word campus so that the vowels always come together?Hence, a total number of words formed during the arrangement of letters of word UNIVERSITY such that all vowels remain together is equals to 60480. How many 4 letter words can be formed using the letters of the word college if each letter has to be distinct?∴ 840 four-letter words can be formed when the repetition of letters is not allowed. How many words can be formed from the letters of the word daughter?The number of words formed from ‘DAUGHTER’ such that all vowels are together is 4320. How many possible arrangements are there for the letters in massasauga in which 4 a/s are together?= 25,200, where 4! is the number of arrangements for the four A’s, 3! for the three S’s, and the 1’s for the rest of the single letters. If the four A’s were put into one group together, then the total number of arrangements for the word MASSASAUGA = 7!/(3! 1! 1! How many arrangements of the letters in Tallahassee has no adjacent A’s?Find the number of arrangements of the letters in TALLAHASSEE which have no adjacent A’s. The number of possible arrangements of the remaining letters is M=8!/(2!) 3. The dashes can be filled in N=C(9,3). How many different strings of length 4 can be formed using a letters of the word flower?×4! Hence, the number of four lettered words can be made with the word FLOWER is 360 words. How do you find arrangements in math?How many different arrangements of the word mathematics are possible where the vowels are always together?In [4989600]distinct ways, the letter of the word ‘Mathematics’ can be written. (i) When vowels are taken together: In the word ‘Mathematics’, we treat the vowels A, E, A, I as one letter. Thus, we have MTHMTCS (AEAI). How many arrangements are possible for word Bahama?They can be arranged in 4! How many different letter arrangements can be made from the letters in the words of River Mississippi count the space too?There we go! There are 34,650 permutations of the word MISSISSIPPI. How many words can be formed by using all letters of the word Delhi?The word ‘DELHI’ contains 5 different letters. Required number of words = Number of arrangements of 5 letters, taken all at a time = 5P5 = 5 ! = (5 *4 *3 *2 *1) = 120. In how many different ways can the letters of the word ‘RUMOUR’ be arranged ? How many arrangements can be made out of the letters of the word Sunday?There are 720 different ways to arrange the 6 letters in SUNDAY. Distinguishable Permutations of Letters in a WordPermutations and Combinations – Arranging Letters Part 1 | Don’t Memorise | GMAT/CAT/Bank PO/SSC CGLPermutations Involving Repeated Symbols – Example 1Permutations: How many ways to rearrange the letters in a word?how many different arrangements of the letters in the word parallel are there? 40,320 56 3,360how many different arrangements of the letters in the word school are there?how many arrangements of 3 digits can be formed from the digits 0 through 9?evaluate c(5, 2).evaluate p(6, 2).evaluate c(7, 7).evaluate c(5, 2). 60 20 10 in how many ways can a gymnastics team of 4 be chosen from 9 gymnasts? |
How many combinations you can create by rearranging the alphabets of the word Indians so that the vowels always come together?
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