[murthymsr]

Originally Posted by Hans_Reiser

Though the solution is extensive its open-ended.
The solution to the problem should be like this
"There cant be more than X(i.e 105 or whatever) roses planted"
The problem I see with your hexagon based solution is that its more a trial and error based solution where you hope that the count is more than the earlier arrangement.
My compliments for reaching 105(which seems the largest), just that the proof isnt generic.

My last response may have to be considered as a sum-up of all the deliberations rather than an answer. (Of course it indicated the answer also.) This would ensure that all participants, as well as any subsequent visitor to this thread, will get the complete picture.

Due to request from some members, for further problems posted by me, Please refer to the thread 'Concepts-total-fundas' at:

http://www.pagalguy.com/forum/quanti...al-fundas.html (Concepts...total fundas!!)

[**genius_yogi**]

help


help


in ineqealities......

[saurabh1310]

cant v plant them in circular way..i guess it'll be more roses then...

[kabir.kukreti]

Hi actually I was seeing modular arithmetic threads can I have a name of any good book on modular arithmetic ?

[murali13_84]

hi i have following doubts pls some1 help me solve this with explanation
a) 1 red 3white and 2blck flags are arranged such that
1. no 2 adjacent flags are of same colour
2. the flags at the end shud b diff color
how many arrangements are posible

[murali13_84]

in how many ways can 4 prizes each havung 1st, 2nd, 3rd position b given to 3 boys if each boy is eligible to recieve more than one prize. pls answer with explanation

[catapult]

Remembered this when I saw the thread topic. I feel it may be relevant.

Comfortably numb..: An ode to a genius: Koteswara Rao

[skrishnak]

i think it will be 6*6*6*6 ways

[ramanujan]

Originally Posted by murali13_84

hi i have following doubts pls some1 help me solve this with explanation
a) 1 red 3white and 2blck flags are arranged such that
1. no 2 adjacent flags are of same colour
2. the flags at the end shud b diff color
how many arrangements are posible

see 3 W means all W can't be in middle ..
we can have

W_W_W_

or

_W_W_W


now 1st way..
3 blank places can be filled in 3 ways 3!/2!1! = 3 (bcoz 2 identical b and 1 r)

WRWBWB WBWRWB WBWBWR

and in the second case

similarly again we have 3 cases

RWBWBW BWRWBW BWBWRW


so we have 6 different cases

[murali13_84]

thanks for your explanation.. i understood the problem now