21coz i stil feel permutatn is necessary
A bag contains unlimited number of white, red, black and blue balls. The number of ways of selecting 10 balls so that there is atleast one ball of each colour is......
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26 Answers
21is my ans FINALLY crrct or abhi bhi kuch changes chahiye?? [7]
1rite got it now...wen dey ask such questions in d brds or JEE, i'll think of another question nd den solve...hehehe. But yep, i got d idea now. thx
21yah! dat was careless i know we gotta put a factor if only 2 only 1 are selected outta 3.....
4^10 - (4C1) * (3^10 - 2^10 - 1) - 4C2 * (2^10 - 1) - 4
62but for that the answer should be
210-2 isnt it
first we chose balls in whatever way we want
then we substract the cases when we have chosen balls of the same color.. there are two such cases.. either all red or all blue... so the answer.!
62you are right ritika in a way.. but then if you are saying this then it means that we are dealing with the case when all red balls are similar!
62yes ritika and tapan...
You guys are ignoring possibility when all the 4 colors were not the first ones to be picked..
take a simpler question..
there are 2 colored balls.
YOu have to pick 10.
what wud your answer have been?
62This is another one of inclusion exclusion principle
can you think this out abhirup?
21select first 3 bals of 1 color each........ permute them......... 3!
rest can be chosen rfm ne of da colours............. 3^7
ans : 3!*3^7
1Assume that you have 4 balls of a color each. Then the remaining 6 balls can be of any color i.e. 4 x4 x4 x4 x4 x4 = 4^6 Is this logic lacking somewhere? Coz all my reasoning in p&c is mostly wrong :(
62This question can be done in two ways.
One when all balls of one color are identical.. another when they are not not identical...
When they are identical it is same as solving
a+b+c+d =10
x+y+z+w=6
= 9C3
in the other case, you have to apply Inclusion exclusion principle.
what is the logic of your answers tapan and ritika?
