[RESOLVED] probability question

[Harsh Gupta]

hi all!!

another one!!

"n" couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is ???

the options are: 3n, (2n)!, 2n ..............sorry there shud be 4 options available but i can see only 3 of them.............but hope u all will get the idea!!

thnx!!

[The_Duck]

I am not sure I understand this but how about simply n couples?

[zaza]

Hm, can you really not work this out? Not even by just writing the possible combinations down?

B = Both
W = Wife only

Try two couples:

Possibilities=4:

B, B
B, W
W, B
W, W

Three couples:

Possibilities=8:

B, B, B
B, B, W
B, W, B
B, W, W
W, B, B
W, B, W
W, W, B
W, W, W

Now try it for 4 couples, 5 (etc) until you can guess the answer.

[Rassis]

If I understood the question correctly, the answer is quite straightforward: n^2

As any one couple is composed by an husband Yi and a wife Xi, then for:

Two couples: 2^2 = 4, that is, (X1+Y1), (X1+Y2), (X2+Y1), (X2+Y2);
Three couples: 3^2 = 9, that is, (X1+Y1), (X1+Y2), (X1+Y3), (X2+Y1), (X2+Y2), (X2+Y3), (X3+Y1), (X3+Y2), (X3+Y3);
Four couples: 4^2 = 16, that is, (X1+Y1), ..., (X4+Y4)
and so forth...

[zaza]

I don't think you have understood the question correctly. Or maybe I haven't. I understand it to mean:

There are n invitations to the party. Each invitation could be taken by either the Wife only, or the Husband and Wife together. Not by the Husband alone. How many possible combinations are there for the uptake of the invitations?

Hence my previous answer.

zaza

[Rassis]

Zaza,

I have read the question once more and I think you are right - I missunderstood it. I agree with your answers for the two and three couple cases examples. Therefore, the number of different gatherings possible at the party is given by 2^n. In the case, for instance, of four couples, you would get the table underneath - remember binary logic?

Thanks

http://img369.imageshack.us/img369/5645/zaza1ex.jpg (http://imageshack.us)

[Harsh Gupta]

Hm, can you really not work this out? Not even by just writing the possible combinations down?
well i thought the same thing..............but i usually make mistakes in Probability questions!!! :(

thnx zaza n Rassis...............yes it shud be 2^n!!!!

PS - plz help with the other question i posted!!! (again probability :( )