solve this problem

[sorabh]

I was wrong.

I said that rearranging the colours on faces will not make any difference. But this is incorrect. here again the adjacent/ opposite thing will have to be considered.



ByeHi Keviv,
Consider this.
we open up the cube... and the six faces are like this.
__5
1_2_3_4
__6

now first i thought 6 different colors can be arranged in 6!
ways..as every combination is a new one.
but that holds in a linear arrangement as you know..because
6! doesnt hold in a circular arrangement.
i know there are other ways to unfold this cube..
but in nocase there would be more than 4 consecutive sections
and the other two can be on either side or on same side.makes no difference as they can be arranged in only 2! ways among themselves.

i think
6c4 * 4! * 2!

6c4 to pick any 4 colors first and put them in a line and then 4! to
arrange them among themselves and 2! for the rest of the two left
out colors. This comes out to be 720..!!!!!
but then... isnt 6!=720...???
so does 6! will hold good.
this is my take what do u say guyz. ?

[keviv]

Originally Posted by sorabh

]


i think
6c4 * 4! * 2!

6c4 to pick any 4 colors first and put them in a line and then 4! to
arrange them among themselves and 2! for the rest of the two left
out colors.

Hi,

the approach you have used is incorrect.

The main glitch arises due to bringing in of the 6C4 thing. It means that you are not considering the identical nature of all sides of the cube.

Take an example,

according to you, arragements

(1) colours a,b on top and bottom ( 5 & 6 in ur diagram) and c,d,e,f on the side faces (1,2,3,4 in ur diagram) is different from (2) c,e on top and botton and a,d,b,f on side faces. But when you look from sides (2) is same as (1).

( I know you do not mean only top and bottom by 5 & 6. But it is definitely one of the implications and it does violate what you have said)

4! is not correct because a,b,c,d and d,a,b,c becomes one and the same arrangement.

2! is not correct because a on top and b on bottom is same as a on bottom and b on top.

The question is about how to colour the cube and not to how to place it or look at it.

Bye.

[cattie]

Originally Posted by keviv

I was wrong.

I said that rearranging the colours on faces will not make any difference. But this is incorrect. here again the adjacent/ opposite thing will have to be considered.

So for a particular choice of colours 'a' and 'b' on opposite faces , the colours on remaining 4 faces can be arranged in 3 ways. Say the colurs are c, d, e, f. in one c,d will be opposite, in the second, c,e and in the third, c,f. This will also take care of the arrangement on the remaining pair of opposite faces.

So total no. of ways = 5 * 3 =15.

thus, i second Sachin's answer.

Correct me if I am still wrong.


Bye.

sorry for this late intrusion but I'd like to make a minor change to the above sol.

Everythings fine as far as placing the pairs c-d c-e and c-f is concerned but then it does matter the manner in which u place the other two colors. It's not the same if u interchange them among themselves, the combination would be different clockwise and counter-clockwise.

Ss the number of combinations possible is 5*3*2 = 30 ways.

[cattie]

no replies yet????

wht do u guys think, keviv, saurav, deepu and the rest?

anyways rajbahadur tomar is backing me :mg;

[roopesh_jindal]

hello ppl
I am a newcomer on this forum
have a ques for u all

Ques : what is the probability that the length of a chord of a circle will be lesser than or equal to the radius of the circle .

I am also giving the answer but not one, but two answers
You ppl tell me which approach is right or if both are wrong,give me the correct approach.

Ans 1):
Consider the angle subtended by the chord of length R at the centre of the circle
It will be 60’ ( quite obviously)
The angle can vary from 0’(a point ) to 180’( diameter)
Hence the probability that a chord will will be lesser than or equal to the radius of the circle is given by (60’-0’)/(180-0) = 0.333333 = 33.33%

Ans 2)
Consider the height of the chord of length R at the centre of the circle.
Since an equilateral triangle is formed among the chord & two radii hence the altitude of the triangle (considering the chord as base ) is R sin 60’= .866R

The height of the triangle can vary from 0 (ie diameter) to R (ie point on circumference)
Favorable cases are from 0.866R( chord of length R) to R (ie point on circumference)
Hence the probability that a chord will will be lesser than or equal to the radius of the circle is given by (R-0.866R)/(R-0) = 0.134 = 13.4%

So which is the correct answer 33.33% or 13.4%

[reachmonil]

Hie!!

though i m a bit late for the reply.... i still felt i could add another dimension to the topic... with my logic!!

neways.... here we go

6 identical sides. lets fix one side.... n paint it with color C1.

the opp. side can be painted in 5 different ways.

now the remaining 4 unpainted sides form a circle and can be painted in (4-1)! ways. = 6.

hence my answer - 5*6 = 30.

wat say folks?!?

[GauravShah]

Originally Posted by roopesh_jindal

hello ppl
I am a newcomer on this forum
have a ques for u all

Ques : what is the probability that the length of a chord of a circle will be lesser than or equal to the radius of the circle .

I am also giving the answer but not one, but two answers
You ppl tell me which approach is right or if both are wrong,give me the correct approach.

Ans 1):
Consider the angle subtended by the chord of length R at the centre of the circle
It will be 60’ ( quite obviously)
The angle can vary from 0’(a point ) to 180’( diameter)
Hence the probability that a chord will will be lesser than or equal to the radius of the circle is given by (60’-0’)/(180-0) = 0.333333 = 33.33%

Ans 2)
Consider the height of the chord of length R at the centre of the circle.
Since an equilateral triangle is formed among the chord & two radii hence the altitude of the triangle (considering the chord as base ) is R sin 60’= .866R

The height of the triangle can vary from 0 (ie diameter) to R (ie point on circumference)
Favorable cases are from 0.866R( chord of length R) to R (ie point on circumference)
Hence the probability that a chord will will be lesser than or equal to the radius of the circle is given by (R-0.866R)/(R-0) = 0.134 = 13.4%

So which is the correct answer 33.33% or 13.4%

Good Q

I think the second approach is the correct one and hence the naswer will be 13.4 %

As for the 1st option, here is y i feel there is an error

Hypothetically each 1 mm of distance on the height shud contain X no. of chords, which shud be equal throughout the height, But for every 1mm, the angles doesn't vary linearly.

Difficult to explain basically, please someone help

Gaurav.

[keviv]

Yup Cattie, reachmonil,

I too feel that 30 is ideed the correct answwer.

So rajbaahadur tomar, what say? Lock kar diya jaaye? Saahi jawab?

[cattie]

Good ques Gaurav,

been bogging myself since yes'day with it , the only thing that I've been able to come up with is yet another seemingly possible sol with yet another diff ans

ans3) The possible lengths of the chords ranges frm 0-2R and the favourable cases are frm 0-R.
probab= (R-0)/(2R-0)= 1/2.

might seem too stupid or simple but then, why not.

[supervish]

No, again its not necessarily linear.

the perpendicular from the centre onto the chord idea is right, it is a linear variance then. Think abt it.

In the other approaches, ur taking a range, wherein the chord variation is not linear.