33q 4 doubt tha usme [7] abhi dale [3]
Some questions on Permutation and combination for revision.....
Q.1 Find the no of ways in which a 3 digit number xyz can be formed from the digits {1,2,3,4,5,6,7,8,9} such that x y z are in AP and x≠y≠z.
Q.2 The number of numbers of the form 7m+7n where m,n belongs to {1,2,3,4,....,99,100} which are divisible by 5.
Q.3 Seven people goes to a cinema hall and park their cars outside which are identical. While returning only 3 of them are able to open their cars(assume no one is a thief i.e. one can open a car by its key only). Find the total no of ways in which it can be done.
Q.4 [7] The no 3 can be written as 3,2+1,1+2 or 1+1+1 in four ways. In how many ways the number n can be written.
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19 Answers
33:P
actually isse prabhavit the : how to get ur paper published :P
http://targetiit.com/iit_jee_forum/posts/how_to_get_your_paper_published_208.html
[3]
62isnt it essentially the same proof as mine :)
just din want to jargonize it :P
;)
33In q 1
I got a very interesting equation
x+1≤y≤[(9+x)/2]
x,y belongs to {1,2,3,..,9}
there fore answer is
x=7
Σ n(y)(x)
x=1
and i solved it :D
62MAQ, your explanation is wrong because you have not considered the cases when you could write say 5 as
2+2+1
in your method only 1 number of the sum is allowed to be non-1
62oops.. ok :D
I din realise that :D :P
Yeah and that one is 2n-1
:)
62Q.1 Find the no of ways in which a 3 digit number xyz can be formed from the digits {1,2,3,4,5,6,7,8,9} such that x y z are in AP and x≠y≠z.
The solution is relatively simple if you start by picking y first!
when y=1 no soln
y=2, 2 solutions
y=3, 4 solution
y=4, 6 solutions
y=5, 8 solutions
y=6, 6 solutions
y=7, 4 solutions
y=8, 2 solutions
y=9, 0 soutions..
now this must be obvious?
33@ Nishant Bhaiya..
it should be 2n-1 naa?
tukka and got it correct sayad.. but don know the method..
and they were not my doubt.. so why u solved ... :(
Jo doubt rahta hai na usme [7] dal dete hai...
13yeah... u r rite bro... [1]
P.S. :
btw... who is mak here...? i donno any mak... [3] [4] ... i'm maq... [6]
13for Q.no.4 ...
n = 1+1+1+...(n times)
now adding any two consecutive 1's is n-1 ways...
adding any three consecutive 1's is n-2 ways...
.
.
. adding all 1's is 1 way...
hence all possible cases can be summed up as 1+2+3+...+(n-1)+1 ..... last '1' is the case where no consecutive 1's are added...
=> no. of ways n can be written in the given format = n(n+1)/2 .....
let me know if my interpretation is wrong...
62yes mak you are right.. but the question says how many unique numbers... so i guess mine is the more correct solutino... but anyways the important thing here is to understand the method :)
13@nishant bhaiya...
in second question... are m,n identical or different...?
if they are identical, then ur answer is correct...
but if they are considered as different.... then for each case, two new cases arise... lyk...
n=4k+1,m=4k+3 and n=4k+3,m=4k+1...
n=4k,m=4k+2 and n=4k+2,m=4k...
then the answer would be 2(25.25+25.25)...
62The no 3 can be written as 3,2+1,1+2 or 1+1+1 in four ways. In how many ways the number n can be written.
answer should be 2n
I have some justification.. but not the full justification.... So i din post the justification...
It should be simple but just not clicking at the moment!
62Seven people goes to a cinema hall and park their cars outside which are identical. While returning only 3 of them are able to open their cars(assume no one is a thief i.e. one can open a car by its key only). Find the total no of ways in which it can be done.
no of ways of choosing the 3 ppl = 7C3
no of ways in which the remaining 4 cant pick their car correctly
??? (This is a standard one) (Isnt it?)
62Q.2 The number of numbers of the form 7m+7n where m,n belongs to {1,2,3,4,....,99,100} which are divisible by 5.
7 -> 2
72 -> 4
73 -> 3
74 -> 1
75 -> 2
then it is cyclic with the remainders....
So for every time power is 4k+1 the remainder will be 2 so we need the m to be such that it has remainder 3 .. hence m should be of the kind 4k+3
similarly if n is 4k+2 , m will be of the form 4k+4
Now this solved already...
i think it should be 25.25+ 25.25