1no....not about that,,,,,,i got the previous one....the 2nd question
a 5 letter word is to be prepared such that the letters appearing in the odd numbered positions are taken from the letters which appear without repitation in the word "MATHEMATICS" and letters appearing in the even positions are taken from letters which appear with repitation in the same word"MATHEMATICS".the number of ways in which 5 letter word can be formed is????
How do we solve circular permutations sums ....i mean for questions like no boys are together??
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21 Answers
62gr8.. i thought it wud not have been enuf...
do let me know if u felt the same. i ll try to post some more !
62gr8.. i thought it wud not have been enuf...
do let me know if u felt the same. i wud try to post some more !
62I know that this article is not complete.. but am already too sleepy. Will try to post some tommorow!
62The two arrangements above are considered to be the same in a circular seating problem.
The two arrangements above are considered to be the same in a circular seating problem.
So in such problems, we always look at keeping one person fixed.
Then, working with the other people.
The important and crucial point a lot of times is to choose whom to fix. !! This is the most important step in solving such problems
Simplest case:
N People, N seats. Round table. Find number of ways!
Soln) Fix one person. Give him a seat (1 way) You have to first fix the person to a seat. (otherwise, it would have been n ways!)
Now, we have N-1 people and N-1 seats. No of ways is (N-1)!
Simple Case:
N boys, M girls, No of ways.
Soln) Fix a boy to a seat. Now we have M+N-1 Seats, N-1 boys, M girls remaining.
Number of ways to arrange them? (N+M-1)!
Why is it not (N+M-1)! / (n-1)!M! ???
(because the boys and girls are all unique. So, in some sense It is the same as the simple case!)
Simple Case:
N boys, M girls, No of ways such that no two boys are together! M>=N
Soln: The motivation is that we have to put boys between girls!
We will have to put girls first. So we need that the person to be fixed has to be a girl.
First, arrange the girls. No of ways to do that is (M-1)! Now there are M spaces between the M girls.
We have to place N boys in these M spaces. No of ways to do that is MPN
Hence, the total no of ways to do this is (M-1)!. MPN
Average case:
N boys and girls. Mahesh and Swati need to sit together.
Soln: Fix Mahesh. Swati can sit in 2 ways. To his left or right. Other N-2 ppl will have N-2 seats.
Total no of ways = 2. (N-2)!
The Key thing is to figure whom to fix!
62ooops... sorry u will get it. havent started to write.. got busy with some other work on the site :)
how about Tommorow morning?
I mean i dont want to give a small stupid answer for a good question of this kind!
62hmm... i dont know if i understood the question properly..
if u tell me the answer.. may be i can give u some idea....
Cos if i understood it properly, i have given the right answer..
the qeustion to be true is a bit vaguely framed! (or may be my english!)
62ur 2nd part i will reply in a larger article... may be today night?
will that be fine?
with a couple of examples?
62ok.. just check and see if the answer matches the solution...
may be i goofed up somewhere else... in trying to be quick..
62odd numbered positions are taken from the letters which appear without repitation in the word "MATHEMATICS"
these are the only letters which appear without repition..
MAT are repeated...
HCEIS are not!
62if repition is allowed..
evens can have only M or A or T
so there are 3 options..
there are 2 places to fill..
so it will be 3*3 !
62give me sometime for the 2nd part.. i will give an article or a good long explanation!
1no....the 2 nd part of the 1st ques is not right i gez.....(in my opininon)
you are chosing 2 different letters.....since it is with repitation,i can 2 Ms or 2A's ...and many more combinations ...isnt it??
and regarding the 2nd one....
plz help me with it...im finding trouble there...
623 odds to be chosen from HEICS (5)
2 evens to be chosen from MAT(3)
No of ways to arrange these 3!.2!
no of ways to chose odds= 53 (if repetition is allowed)
no of ways to chose evens = 32 (if repetition is allowed)
multiply these 3...
does it make sense?
623 odds to be chosen from HEICS (5)
2 evens to be chosen from MAT(3)
No of ways to arrange these 3!.2!
no of ways to chose odds= 5c3 (if no repetition)
no of ways to chose evens = 3c2 (if no repetition)
multiply these 3...
does it make sense?