1
Bicchuram Aveek
·2009-11-11 09:04:09
2nd part ))) 2 balls total in 2 boxes or 2 balls each in the 2 boxes ?????
1
shreya
·2009-11-11 09:32:42
there are only 5 balls
u got to place only 1 ball in each box
49
Subhomoy Bakshi
·2009-11-11 10:20:18
first one is damn easy.......
one way is there to put balls in respectively same coloured boxes....
total way to put coloured balls in the boxes is 5*4*3*2*1=120
so reqd. no of ways=120-1=119
49
Subhomoy Bakshi
·2009-11-11 10:30:37
lolz......second one is easier....first we choose 2 boxes out of 5 in 5C2ways.these two boxes can be filled in 1 way each.....others can b filled in as 3*2*1 ways....
so total arrangements=5C2*3*2*1 =60 ways....
66
kaymant
·2009-11-11 18:04:58
@subhomoy, both of your results (in #4 and #5) are wrong.
49
Subhomoy Bakshi
·2009-11-11 20:16:02
was my logic right?????????????
24
eureka123
·2009-11-11 20:27:19
1)5![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}]
2)Atleast 2 balls in same color
=> 3 balls in not their box +2 balls in not their box +1 balls in not their box + All balls in their boxes
=>^5C_2.3![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}]+^5C_3.2![1-\frac{1}{1!}+\frac{1}{2!}]+^5C_4.1![1-\frac{1}{1!}]+1
49
Subhomoy Bakshi
·2009-11-11 21:11:51
oops.....nvr saw the at least one......bt wheres the mistake in the frst 1??????
1
shreya
·2009-11-12 02:21:56
@eureka , how did u think this logic for ques. (b).????
24
eureka123
·2009-11-12 02:32:18
@subo..its simple dearrangements in case1
@shreya...didnt get ur dbt....[7]
1
shreya
·2009-11-13 09:16:49
@ eureka,
actually i have plenty of such questions...
i know when to apply dearrangement but i am never able to think the logic for the part B of the question.
like here, how did u did
"Atleast 2 balls in same color
=> 3 balls in not their box +2 balls in not their box +1 balls in not their box + All balls in their boxes"
is there any general approach for this...??!!
24
eureka123
·2009-11-13 09:19:38
ya its gneeral approach..see ques said atleast 2 balls of same colr..u can coorelate this with sputting lettersi n correct envelops..
so when we need min.2 balls in correct box...that means 2 box in same color +3 balls in same color +4 balls in same color + 5 balls in smae box
so converse of this will eb to simplify the calculation
3 balls in not their box +2 balls in not their box +1 balls in not their box + All balls in their boxes
I hope u r getting rite ans using them
1
Arka Halder
·2009-11-13 09:30:05
yeah.first one is given by formula n![(1/2!)-(1/31)+...+(-1)n/n!]
answer for 1st is 32.
as suggested by eureka earlier,we can use this formula in the 2nd case to find the answer.
66
kaymant
·2009-11-13 10:54:46
for all of you:
its derangement and not dearrangement.