#1
7th September 2015, 04:58 PM
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Permutation and Combination for IBPS PO
Hello friends , I want to appear in PO Bank exam Which is Conducted by IBPS , for it Will you please provide me Permutation And Combination Section Question paper for my help ??
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#2
7th September 2015, 05:00 PM
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Re: Permutation and Combination for IBPS PO
Permutation And Combination is primarily because various questions from this section tests candidate's analytical skill. This topic just involves basic calculations. Permutation implies arrangement where order of things is important and includes word formation, number formation, circular permutation etc. Combination means selection where order is not important and it involves selection of team, forming geometrical figures, distribution of things etc. Factorial = Factorial are defined for natural numbers, not for negative numbers. n! = n.(n-1).(n-2).........3.2.1 for example: 1) 4! = 4.3.2.1 = 24 2) 6!/ 4! = (6.5.4!)/ 4! = 6.5 = 30 3) 0! = 1 Problems for practice: Problem 1: Choose permutation or combination 1) Selection of captain and bowler for a play. Permutation 2) Selection of four students for a lecture. Combination 3) Assigning people to their seats during conference. Permutation Problem 2: Evaluate 7P2 . 4P3 Solution: (7!/ 5!). (4!/ 1!) ⇒(7.6). (4.3.2) ⇒1008 Problem 3: Evaluate 5C2. 3C2 Solution: (5!/3!2!). (3!/2!1!) ⇒(5.4/2). (3) ⇒30 Problem 4: How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females? Solution: This is a case of combination i.e.selecting 3 males from 6 males and 2 females from 5 females. ⇒Required number of ways = (6C3 *5C2) ⇒(6.5.4/3.2)*(5.4/2) ⇒200. Problem 5: How many words can be formed by using letters of the word "DAUGHTER" so that the vowels come together? Solution: This is a case of permutation. in a word "DAUGHTER", there are 8 letters including 3 vowels (AUE) According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels) These 6 words can be arranged in 6P6 ways ⇒6P6 = 6!/1! = 6! = 720 WAYS Also, three vowels in a group may be arranged in 3! ways ⇒3! = 6 ways Therefore, required number of words = (720*6) = 4320. Take a Screen Shot of this page : Here I am Sharing some no. of Question of Permutation And Combination for your help : Question 1 In how many different ways can the letters of the word "CHARGES" be arranged in such a way that the vowels always come together? a) 1440 b) 720 c) 360 d) 240 Answer : a) 1440 Solution : The arrangement is made in such a way that the vowels always come together. i.e., "CHRGS(AE)". Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways. The vowels "AE" can be arranged themselves in 2! ways; i.e.,2! = 2 ways Therefore, required number of ways = 720 x 2 = 1440 ways. Question 2 In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions? a) 1440 b) 43200 c) 1440 d) 5420 Answer : b) 43200 Solution : There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants. Let us mark these positions as under: [1] [2] [3] [4] [5] [6] [7] [8] [9] Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9. Number of ways of arranging the vowels = 5P3 = 5x4x3 = 60 ways. Also, the 6 consonants at the remaining positions may be arranged in 6P6 ways = 6! ways = 720 ways. Therefore, required number of ways = 60 x 720 = 43200 ways. Question 3 In how many different ways can the letters of the word "CANDIDATE" be arranged in such a way that the vowels always come together? a) 4320 b) 1440 c) 720 d) 840 Answer : a) 4320 Solution : There are 9 letters in the given word, out of which 4 are vowels. In the word "CANDIDATE" we treat the vowels "AIAE" as one letter. Thus, we have CNDDT(AIAE). Now, we have to arrange 6 letters, out of which D occurs twice. Therefore, number of ways of arranging these letters = 6!/2! = 720 / 2 = 360 ways. Now, AIAE has 4 letters, in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4!/2! = 1 x 2 x 3 x 4 / 2 = 12 Therefore, required number of words = (360 x 12) = 4320. Question 4 In how many different ways can the letters of the word "RADIUS" be arranged in such a way that the vowels occupy only the odd positions? a) 72 b) 144 c) 532 d) 36 Answer : d) 36 Solution : There are 6 different 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 3 marked 1, 3 and 5. Number of ways of arranging the vowels = 3P3 = 3! = 6 ways. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6 ways. Therefore, total number of ways = 6 x 6 = 36. What will be the largest number when the position of the first and the second digit within each of the given numbers are interchanged? a) 549 b) 369 c) 827 d) 453 e) 968 Answer : e) 968 Solution : Given numbers: 549, 369, 827, 453, 968 After interchanging the first and second digit, the numbers are: 549 -> 459 369 -> 639 827 -> 287 453 -> 543 968 -> 698 Therefore, the highest number is 698 and its corresponding number is 968. Hence the answer is option e. What will be the sum of the first and last digit of the smallest number if the position of the first and last digit within each of the given number is interchanged? a) 14 b) 12 c) 15 d) 7 e) 17 Answer : d) 7 Solution : Numbers after interchanging first and last digit are, 549 -> 945 369 -> 963 827 -> 728 453 -> 354 968 -> 869 Now the smallest number is 354 and the sum of first and last digit is 3 + 4 = 7. Submit Your Resume for Bank & Finance Jobs Which of the following will be the second digit of the second lowest number when 7 is added to each of the given five numbers? a) 6 b) 7 c) 3 d) 8 e)5 Answer : a) 6 Solution: After adding 7 to each of them, they become: 549 + 7 = 556 369 + 7 = 376 827 + 7 = 834 453 + 7 = 460 968 + 7 = 975 Second lowest number is 460. And required second digit is 6. Find the last digit of the middle number, when 3 subtracted from the first digit and added to the second digit of each of the given five numbers. a) 7 b)9 c) 1 d) 6 e) 3 Answer : b)9 Solution: Given numbers: 549, 369, 827, 453, 968 After adding 3 to the second digit and subtracting 3 from first digit, they become: 549 -> 279 369 -> 099 827 -> 557 453 -> 183 968 -> 698 In ascending order: 99, 183, 279, 557, 698 Therefore, the middle number is 279. Last digit of the middle number is 9. |