How many combinations can be formed from the letters of the word hexagon taken 3 at a time?
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Introduction:-Permutation & combination deal with the techniques of counting without direct listing of the number of elements in a particular set or the number of outcomes of a particular experiment. The content of this article may be too rudimentary for most readers, but for beginners, it will be helpful.one has to work hard on basics. Once the basics are very clear, permutation is a very systematic subject. Case 1:- “MY FRUIT SALAD IS A COMBINATION OF APPLES, GRAPES & BANANAS”. We don’t care what order the fruits are in, they could be “grapes, apples & bananas” or “bananas, grapes & apples” it’s the same fruit salad. Case 2:- “THE COMBINATION TO THE SAFE LOCK IS 289”.Now we do care about the order. “982” would not work nor would “892”. It has to be exactly 2-8-9. In mathematics we use more accurate language:
A Permutation is an ordered combination. PERMUTATIONS:- The number of permutations on ‘n’ different things
taken ‘r’ at a time is the same as different ways in which ‘r’ places can be filled up with ‘n’ given things. It is denoted by npr or p (n, r) & is given by Type 1-When a certain number of things always occur together 1) In how many ways can 7 people be seated in a row for a photograph, if two particular people always want to be together. Solution:-Consider the two people who want to be together as 1
unit. The number of permutations of remaining 5 people + 1 unit (2 people) = six units is 6! Type 2-When certain number of things occurring together is more than one type Solution:-Consider all 5 quant books as 1 unit, 4 reasoning books as 1 unit & 6 English books as 1 unit. So, total 3 units can be arranged in 3! Type 3-When certain things of one type do not occur together Solution:-Total ways in which best & worst are never together = Total arrangements – Two papers are always together Type 4-Formation of numbers with digits (digits can be repeated) Solution:-Three digit number consists of 1unit place, 1 tens place &1 hundred place. In the given problem there are 5 numabers, as it can be repeated any no.of times. COMBINATIONType 5-When no two of a certain type occur together Solution:- If no two girls can be together, we first arrange the boys. This can be done in 9! Ways. In the gaps we have to arrange the girls. Considering the left most & right most positions & the gaps in between the boys. Let ‘B’ be the position of boys & ‘G’ be the position of girls. Type 6-Formation of numbers with digits (digits are not repeated) Case1:- When there is no restriction Solution:- Th H T U The thousand place (Th) can be filled in 7 ways. For each of these the hundred, tenth, units place can be filled in 6, 5, 4 ways respectively Case 2:- Numbers divisible by a particular number or even number or odd numbers Solution:- If the number is even it has to end with even number (2, 4 or 6 in this given question).Hence units place has 3 choices & thousandth , hundredth & tenth place can be filled in 5 ,4,3 ways respectively. Type 7-Word building (Alphabets not repeated) Solution:- The word MONKEY has 6 letters. The six different letters can be arranged in 6P6 = 6! = 720 ways. Case 2:-When few letters/vowels/consonants occur together Solution:- The word LAUGHTER has 3 vowels & 5 consonants. Consider the 3 vowels as 1unit.Hence 1unit+ 5 consonants. Case 3:- When vowels/consonants occupy odd/even places Solution:- The word HEXAGON has 7 letters of which 3 are vowels. Since total letters is 7, there are 4 odd places & 3 even places. Type 8-Permutation of like
things The number of permutations of ‘n’ things taken all at a time, where ‘x’ of the things are alike & of one kind, ‘y’ others are alike of another kind, ‘z’ others are like & of another kind & so on is Solution:- The word ENGINEERING has 11 letters out of which E is repeated thrice, N is repeated thrice, G is repeated twice, I is repeated twice. b) All 3 E’s are together EEE & NGINRING c) No two vowels are together Type 9-Circular Permutations When objects are arranged along a closed
curve like a circle or ring, the permutations are known as circular permutations. Example:- Solution:- In case of necklace there is no distinction b/w the clockwise & anti-clockwise arrangement. Hence no. of ways = (10-1)! /2 = 9! /2 Exercise problems for practice1) Find the value of b)
8P6 Solutions:-1) a) 11880 b) 20160 How many different ways can the letters of the word Hexagon be arranged?question_answer Answers(1)
ways = 5040 ways.
How many different seven letter arrangements of the letters in the word hexagon can be made if each letter is used only once?
How many ways can the letters of the word Heptagon be permuted so that the vowels are never separated?P(7,7) = = = = 5040. They can be permuted in P (7,7) = 5040 ways. The vowels in the word are E, A, O.
How many words meaningful or meaningless can be formed with the letters of the word hexagon?0 ! x 3! = 1201 x 6 = 720. Hence total number of arrangements possible is 720.
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