Permutations combinations and probablity

[mathmoron]

Hi all
I have some problems on PC and Probs that really beats me.....
1. In how many ways can six faces of a cube be painted with 6 different colours ?
-I don't understand how this is a case of circular permutation.....could someone post me the solution with the logic ??

2.Amit went to the market to buy 18 fruits in all.If there were mangoes,bananas,apples and oranges for sale then in how many ways could amit buy the fruits ?
i)18C3 ii)18C4 iii) 21C3 iv) 21C4
-What is the formula thats being used here ? how is it derived ?

3.4 balls are to be put in five boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and balls are similar
d)both boxes and balls are different
-Can someone give me a lucid explanation for these cases ?

ALL THE BEST FOR CAT

-mathmoron

[zango]

Originally Posted by mathmoron

Hi all
I have some problems on PC and Probs that really beats me.....
1. In how many ways can six faces of a cube be painted with 6 different colours ?
-I don't understand how this is a case of circular permutation.....could someone post me the solution with the logic ??

For understanding this concept, you have to understand what a circular permutation really is.
Assume you wish to arrange 3 numbers in a circle. If there was no restriction on the arrangement being circular, the number of ways of arrangements would be 3! = 6. However, since the arrangement is supposed to be circular, therefore, certain arrangements, when rotated, form other arrangements, which have already been counted.
Hence, the arrangement
1
2 3
is same as the arrangement
2
3 1
(simple clockwise rotation).

Now, if we were to fix the positions for one of these elements, then, there would be no confusion and the places would be well defined. This leads us to the formula: n objects arranged in a circle - number of permutations - (n-1)!

Now to come to your question: For a cube to be colored with 6 different colors on its 6 sides. Given above explanation, it is clear that by rotating the cube, it is possible to arrive at a different permutation - which can be considered identical.

However, if you were to take one side as fixed, color it - then you would have 5 other sides relative to this side, on which to put 5 colors.

(well that is my logic - i am not sure if this is correct - i would never have figured this was a circular permutation if your question explicitly didn't say so :P)

all the best.

[mathmoron]

The answer is 30.
From your suggestion i proceed as follows...
1.paint the top face of the cube with 1 colour, the lateral faces can be painted in 3! ways.The bottom face is painted with the remaining colour .But then Im unable to proceed further..Please help me out .Also check the other questions ....thanx

[zango]

Originally Posted by mathmoron

The answer is 30.
From your suggestion i proceed as follows...
1.paint the top face of the cube with 1 colour, the lateral faces can be painted in 3! ways.The bottom face is painted with the remaining colour .But then Im unable to proceed further..Please help me out .Also check the other questions ....thanx

Check this out:
http://www.mathematische-basteleien....%20the%20Cubes

This demonstrates how to arrive at the figure 30.
Apparently - there are 13 axes to a cube - along which if you rotate the cube, you may get 24 sets of similar looking cubes for every possible permutation (6!)
Therefore, 6!/24 = 30.

Ah well, too complicated for me - i would have marked 4! as the answer, at best - worst, i would have gained 1 negative mark (and lost 1 positive mark)

[mathmoron]

Yup , guess this is one of the cases of the problem looking simple and the solution tough.....see if you can attempt the third one .....

[Jyotiraj]

Hi,
Please see my solutions to your questions below.

2.Amit went to the market to buy 18 fruits in all.If there were mangoes,bananas,apples and oranges for sale then in how many ways could amit buy the fruits ?
i)18C3 ii)18C4 iii) 21C3 iv) 21C4
-What is the formula thats being used here ? how is it derived ?


Solution : The formula here is (n+r-1)C(r-1) where n = 18 , r = 4.

Basically, the question is the same as saying, in how many ways can you divided 18 fruits among 4 people? See, think of this way, get those 18 fruits in a line and think about inserting 3 sticks in between them. This will essentially split up the fruits into 4 groups , which is what you want. Now, with 18 fruits and 3 sticks, you have 22 entities, which can be arranged in 22! ways. But we do not bother about the permutaions of the 18 fruits so divide by 18!. we also know that the inserted sticks are identical , so divide by 3! You get 22!/(18!* 3!) which is your answer.

3.4 balls are to be put in five boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and balls are similar
d)both boxes and balls are different
-Can someone give me a lucid explanation for these cases ?


a) Balls are similar. Boxes are different. Same as question above. Ans. 8! / (4! * 4!) = 70 ways.

b) Balls are different. Boxes are different. What you want here are all integral solutions of the equation x+y+z+w+k=4, and then you do combinations with them. For example, one possible solution is (4,0,0,0,0). ie put all 4 balls into 1 box. You can do that in only 1 way because all the boxes are identical, so (4,0,0,0,0) is identical to , say (0,0,0,4,0).
Next another solution is (3,1,0,0,0). Now you need to find in how many ways you can group 3 balls from a total of 4. It;s 4C3= 4. So total no. of ways for (3,1,0,0,0) is 4.
Similarly you need to find out the combinations for solutions of the form (2,2,0,0,0), (2,1,1,0,0),(1,1,1,1,0), They are respectively 6, 6, and 1 way. Total is 1+4+6+6+1=18 ways.

c) if both are similar, you stop at finding the unordered pairs . the answer is 5. They are (4,0,0,0), (3,1,0,0,0),(2,2,0,0,0),(2,1,1,0,0)and (1,1,1,1,0).

d)If everything is different, you need to put everything above together. Which would turn out to be hell. What you should do here is think like this. See, the first ball can go to any of 5 boxes, the second alos to any of 5 boxes , same with third and fourth. so for all balls you have 5 options. So answer is 5^4=625.


Tell me if I am right.

[RahulSharma]

absolutely right buddy

[mathmoron]

thx dude i got the problem......1 correction by the way...that would be 21 entities and not 22 right ?

[Jyotiraj]

Yup....it should be 21 sorry for the typo... there's one more mistake in there, by the way ... i said somewhere that for the solution set of th eform (2,2,0,0,0) you can choose those in 6 ways. It should be actually 3 since the first half of the two objects chosen would get repeated in the second lot of selections. That would bring down that answer from 18 to 15.

Tell me if you don't understand what I am going on about.

[whocarez]

Is there any formula to slove directly for case 2 where balls are different and boxes are similar