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avarma
10-21-2004, 05:05 PM
Find the number of ways in which 5 men and 4 women can be seated round a table so that no two women are together?

avarma
10-21-2004, 05:48 PM
Another one in the same genre

In how many ways can 3 men and 3 women be seated at a round table if there is exactly one person between two particular women?

avarma
10-21-2004, 06:52 PM
Another one
How many different garlands can be made from 6 marigolds and 2 roses?

Dingus
10-21-2004, 09:32 PM
My guesses:
Question1: Seat one woman first, then the remaining 3 need to be seated, this can be done in 3! ways. No two women can be together so the men have to be seated in between. There are five men, so they can be seated in 5! ways.
So finally --> total no. of ways 3! * 5! ways --> 720 ways

Question2: Seat one woman, rest can be seated in 2! ways. Then the 3 men can be seated in 3! ways. So total--> 2!*3! ->12 ways

Question3: There are six marigolds. All are alike. So are the two roses. So this is essentially a permutation of the roses in 8 places.
That is 8P(2) --> 56. Divide by two as they both are roses --> 28

Hope this logic is correct. :crazy:

avarma
10-22-2004, 12:29 AM
Q1: The answer given is 2880

My guess is since no two women can sit together they have to sit in the gaps between men. So first seat five men. This can be done in 4! ways. There are 5 ways to seat the ladies "around" the men. Also the ladies can sit 4! ways among themselves. So 4!*4!*5 = 2880

Q2: Answer given is 48

For the time being leave aside two "particular" women. The remaining 4 persons can be seated in 3! ways. Now these two particular women may be seated "around" any of 4 persons, and further the two can be arranged within themselves in 2 ways.
Hence the required number of arrangements is 2*4 * 3! = 48

Q3: The answer given is 4

I couldn't find any formula to use in this case. But if you go about manually there are only 4 arrangements possible.
Does anybody know of any formula that may be useful in such situations? The number of ways a necklace can be formed from 2 red & 2 blue beads etc....

bbm833
10-22-2004, 07:00 AM
Q1)A suggestion

Answer = total # of ways - #of ways in which two woman are together
=8! - 7! ???

awhig
10-23-2004, 12:47 PM
My guesses:
Question1: Seat one woman first, then the remaining 3 need to be seated, this can be done in 3! ways. No two women can be together so the men have to be seated in between. There are five men, so they can be seated in 5! ways.
So finally --> total no. of ways 3! * 5! ways --> 720 ways

Question2: Seat one woman, rest can be seated in 2! ways. Then the 3 men can be seated in 3! ways. So total--> 2!*3! ->12 ways

Question3: There are six marigolds. All are alike. So are the two roses. So this is essentially a permutation of the roses in 8 places.
That is 8P(2) --> 56. Divide by two as they both are roses --> 28

Hope this logic is correct. :crazy:
Agree with you on Ques3.
In all there are 8 flowers. To arrange them , number of ways = 8!.
Since there are 6 flowers of one kind and 2 of other kind ,
number of ways of arranging reduce to (8! ) / (2! * 6!) = 28.

awhig
10-23-2004, 12:49 PM
Q1: The answer given is 2880

My guess is since no two women can sit together they have to sit in the gaps between men. So first seat five men. This can be done in 4! ways. There are 5 ways to seat the ladies "around" the men. Also the ladies can sit 4! ways among themselves. So 4!*4!*5 = 2880

Q2: Answer given is 48

For the time being leave aside two "particular" women. The remaining 4 persons can be seated in 3! ways. Now these two particular women may be seated "around" any of 4 persons, and further the two can be arranged within themselves in 2 ways.
Hence the required number of arrangements is 2*4 * 3! = 48

Q3: The answer given is 4

I couldn't find any formula to use in this case. But if you go about manually there are only 4 arrangements possible.
Does anybody know of any formula that may be useful in such situations? The number of ways a necklace can be formed from 2 red & 2 blue beads etc....
First one is correct.

buddyboy
10-24-2004, 01:17 AM
for the second question, will the answer be:

Man1 Man2 Man3 Woman1 Woman2 Woman3

if you want to have exactly one person sit particular between Woman1 and Woman2, then you can do,

3 * 3 * 2 * 3 * 2
^ ^ ^ ^ ^
w1 man w2 (m/w) man

so the answer will be 108 for this particular case, again, you have set a particular person sit between Woman 2 and Woman3 or Woman3 and Woman1, anyways, you get:
(108 * 3)/ 3! = 54
correct?

lmtuan
10-24-2004, 06:17 AM
*Answer of question about 5 men and 4 women:
5! * 4! = 2880.

*Answer of question about 3 men and 3 women:
2 * 4 * 3! = 48.
*Suppose that 6 marigolds are similar and 2 roses are similar,too. Then we can be made 4 different garlands.

netwizio
11-19-2004, 07:21 PM
One thing is confusing me a lot in these questions
The persons are to be seated around the table
so is it not the case of cyclic permutation ????

especially in the second one

netwizio
11-19-2004, 08:11 PM
Sorry for impatience i got it

in second question
the answer is 2! * 4 * 3!
2!=> i fixed one woman on one seat ( to get a reference in circular permutation)
then there are two ways to seat the other (the second of the two specific women) either to the left to the right so multiply by two factorial

4=>now there are 4 ways to sit a person between these two women
3!=>obviously the no of ways to sit the rest of the three