word that has the maximum number of repetation of a single alphabet(meaningful one pls)
Answers
Answer Pneumonoultramicroscopicsilicovolcanoconiosis
Explanation: Above answer is from google and it's meaning is lung disease caused by inhaling fine ash and sand dust,
Permutations and combinations is a very interesting chapter and gives us food for thought. Most problems can be solved in multiple ways. Combination of different techniques and our application of ideas is what determines our command over this topic. More over, this topic is very important in learning "Probability" so study the problems with enthusiasm.
It is always confusing for some, when to use permutation and when to use combination.
Combination or selection:
In your college, your college management is planning to form an "anti ragging" committee. There are 3 persons who are interested to be a part of that committee. But the management is thinking of keeping only 2 in the committee. In how many ways this can be achieved?
Our possible combinations are:
AB, BC, AC
Only 3 are possible. We don't consider selection of AB and BA are two different ones. Whether you choose A first and B next or B first and A next may not make any difference.
So remember, When there is a selection like Committee, You have to select r persons/ objects from n persons / objects, we use nCr observe the C in both.
Selection of 3 out of 5 is represented as 5C3which is equal to 5C2 as nCr=nCn−r
How do we write 5C3
Text book notation for nCr=n!r!(n−r)! but we hardly use this formula to calculate 5C3 or any other selection
Simply we do like this. we write 5 in the descending order upto 3 times and divide it by 3!
So 5C3=5×4×33×2×1=10
Or we know that nCr=nCn−r so 5C3=5C5−3=5C2
Now 5C2=5×42×1 = 10
Permutation or Arrangement:
Now, We have to take a photo graph of our newly formed committee and put it in the notice board. How do we do that?
Now 6 different arrangements AB, BA, AC, CA, BC, CB are possible.
Whenever there is an arrangement, we use the formula nPr as like in Photograph where arrangement is possible
So selection of r objects from n objects and arranging them in r positions will be nPr.
Rather than using this formula, you can first select r objects from n, using nCr and multiply it by r! as it arranges r objects in r positions.
Therefore, nPr=nCn−r×r!
Note: When we have to arrange r objects in r positions, this can be done in r! ways. When there are repeatations we have to divide r! with those repetitions.
Solved Examples
(Problems on alphabets)
Solved Example 1:
1. How many arrangements can be made of the letters of the word “ASSASSINATION”? In how many of them are the vowels always together?
Solution:
In the word “ASSASSINATION”, n (total no.of objects) = 13, among them we have four S, three A, two N and two I. So total no.of possible arrangements of these 13 objects = 13!4!×3!×2!×2! = 13!4!×4×3! = 13!4!2
If vowels are always together, we can say that all the vowels taken together (AAAIIO) will be behaving like a single object, than total no.of objects will be 8 (four, S, two N, one T and one group of vowels). Total no.of arrangements for these eight objects will be 8!/(4!x2!). Now the vowels altogether can be arranged in 6!/(3!x2!) ways so the total no.of arrangements of given 13 objects when vowels are always together is 8!×6!4!×3!×2!×2! = 8!×6!4!2
Solved Example 2:
2. In how many ways can the letters of the word ARRANGE be arranged so that
Solution:
By using the following equation we can get the required answer
Two R’s are never together = Total possible arrangements - Two R’s are always together
Total words = 7!2!2! = 7×6×5×4×3×22×2 = 1260
R’s together 6!2! = 6 x 5 x 4 x 3 = 360
R’s never together = 1260 - 360 = 900
(b) The Two A’s are together but not two R’s = Two A’s are always together - (Two R’s and two A’s are always together)
Two A’s together => 6!2! = 360
Also two R’s together = 5! = 120
Ans = 360 - 120 = 240
(c) Neither 2 A’s or 2R’s are together = total possible arrangements - {(two A’s are always together + Two R’s are always together) - Both Two A’s and Two R’s are always together}
Either 2A’s or 2R’s or both A’s & R’s are together = 600
Hence None of these together = 1260 - 600 = 660.