View Discussion Improve Article Save Article Like Article View Discussion Improve Article Save Article Like Article Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter. In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered. Permutation Formula In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.
Combination A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used. Combination Formula In combination r things are picked from a set of n things and where the order of picking does not matter.
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Saurav S. how many permutations can be formed by the letters of the word 'VOWELS' when 1>the letters can be placed anywhere 2>each word begins with E 3>each word begins with O and ends with I 4>all vowels come together 3 Answers By Expert Tutors
Ralph K. answered • 01/05/15 Effective Mathematics, Physics and Chemistry Tutor
1) This is the easiest to do, basic permutations with 6 objects. The answer is 6! or 6*5*4*3*2*1 = 720 2) This constrains the problem to a fixed first letter and permutations of 5 objects. The answer is 5! = 5*4*3*2*1 = 120 3) This constrains the problem to a fixed first and last letter and promotions of 4 objects. The answer is 4! =4*3*2*1 = 24 4) This is a bit more difficult. The problem now needs to be viewed as O and E together as a single unit. Thus we have permutations of 5 objects which as in number 2 above is 120. This is not the final answer as the vowel grouping has 2 possibilities OE or EO, thus we have 2 times as many permutations which give 2 * 120 = 240.
1) since you have to pick all placements, you have 6 for the first space and can't reuse it, 5 for the next, then 4,3,2 and 1 which are all being multiplied which is the definition of 6! = 720. 2) Since I'm specifying the first space must be an E, I have 5 choices for the next, then 4,3,2,1 all being multiplied which is 5! =120. 3) Since two spaces, the first and last, are specified, you're only picking 4 for the middle 4 spaces, so 4! = 24. 4) The easiest way of doing this is by picking a specific permutation such as the first 2 are vowels and then the other 4 are consonants. This specific case would be 2*1*4*3*2*1 = 48. Now the vowels could have been in positions 2 and 3....or 3 and 4...or 4 and 5.....or 5 and 6. This makes 5 possibilities for having the vowels together and each specific possibility has 48 ways of occurring, so your total possibilities is 48*5 = 240. |