66@subhomoy, both of your results (in #4 and #5) are wrong.
ques.
There are 5 boxes of 5 different colors. Also there are 5 balls of colors same as those of the boxes. In how many ways we can place 5 balls in 5 boxes such that
a) all balls are placed in the boxes of colors not same as those of the ball.
b) ( doubt here ) at least 2 balls are placed in boxes of the same color.
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15 Answers
12nd part ))) 2 balls total in 2 boxes or 2 balls each in the 2 boxes ?????
49first 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
49lolz......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....
241)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
49oops.....nvr saw the at least one......bt wheres the mistake in the frst 1??????
24@subo..its simple dearrangements in case1
@shreya...didnt get ur dbt....[7]
1@ 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...??!!
24ya 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
1yeah.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.