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Numerical Ability
Permutation and Combination
there are 4 boxes coloured red yellow green and blue. If 2 boxes are selected, how many combinations are there for atleast one blue or red box to be selected
Read Solution (Total 5)
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- @ RIMMY:
Consider the following generalized solution,it may be helpful to u.
Generalized Solution:
Let there are m boxes being distinctly colored and there colors be CL1,CL2,CL3,....CLm respectively.
Now, let , we have to choose n boxes (n< or,=m) with the condition that at least
one of k boxes(k< or,= n) with colors CL1,CL2,CL3...CLk should be present in our selection.
It should be mentioned all boxes are distinct since each of them differently colored.
Now, clearly n boxes can be chosen from m boxes in mCn ways .
Now, if we remove all the given k boxes(k< or,= n) with colors CL1,CL2,CL3...CLk
from m boxes, then the number of ways one can choose n boxes from the remaining (m-k)boxes =(m-k)Cn.
And in all these selections ,no1 of k boxes(k< or,= n) with colors CL1,CL2,CL3...CLk will be present.
Now, we know n boxes can be chosen from m boxes in mCn ways.
And again from the above discussion we have ,the number of ways one can choose n boxes from m so that no1 of k boxes with colors CL1,CL2,CL3...CLk will be present on our selection = (m-k)Cn
So, number of the remaining selections = mCn- (m-k)Cn
And , clearly for these cases of selection ,at least 1 of k boxes with colors CL1,CL2,CL3...CLk will definitely be present.
APPLICATION OF THE ABOVE THEORY FOR THE GIVEN PROB:
Here m=4 , n=2 ,k=2
So,mCn- (m-k)Cn = 4C2 - 2C2 = (4!/{2!*2!})-(2!/{2!*0!}) =6-1 =5.
So, the answer is 5.
If any1 of u have any kind of confusion ,complaint or any other views regarding the given prob and my posted solution,then plse do share it with us..
Thanks to all..:-):-) - 12 years agoHelpfull: Yes(30) No(2)
- 5 combinations
BR
BY
BG
RY
RG - 12 years agoHelpfull: Yes(11) No(4)
- yes 5 combinations is the right answer but what is the formula for doing these type of questions, in case the question involves large quantities where large no. of combinations can be made.
- 12 years agoHelpfull: Yes(4) No(1)
- The ans is 16
As in the question it is said that at least one box will be blue or read, so if we choose one box is blue then there may be 4 combination and for read also there will be 4,
so 4*4=16
Or 4C1*4C1=16 - 4 years agoHelpfull: Yes(0) No(0)
- Total number of ways of selecting 2 box out of 4 is 4C2 = 6. After, there was 1 way in which blue and red box come (BR). In question, at least one blue and one red ball is asking, so we reduce the number of ways(6) by 1,therefore 6 - 1 = 5. ANSWER : 5 (Total number of ways in which at least 1 blue and 1 red box selected).
- 4 years agoHelpfull: Yes(0) No(0)
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