1this is not a doubt naa.........leave it..........
no time 2 solve.....only see d solution and understand
[2][3][3][3][3][3].post d solution hehe
It is given that :
If 'n' things are arranged in a row, no. of ways in which they can be deranged so that none of them occupies their original place is
n!\left\{1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{(-1)^{n}}{n!} \right\}
From the above result, deduce the expression for
"no. of derangements, when 'n' things are arranged in a row, such that 'r' things donot go to their original places and 'n-r' things go to their original places "
This is not my doubt, so have patience to give the complete solution (or atleast the required steps in brief) for the benefit of everyone. :)
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7 Answers
1
11a tukka
replace n by r in the expression for derangements
because n-r can go only to their original places and the remaining r can be placed anyway
11r!\left\{1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{(-1)^{r}}{r!} \right\}
13i hink subash is rite...as n-r things can go to rite place in only one way....
1subash, u r partially correct. the things aren't identical, so u must multiply with nCn-r. then it'll be done. :)
now, how will u derive the formula i gave in the question ?
1add one more question.
if the no. of ways of deranging 'n' objects (arranged in a row) such that none occupies their original place is 1854. find the no. of ways in which these objects can be deranged so that '[n/2]'([ ] represent greatest integer function) objects do not go to their original places.