Question

HEEEELP!! :( there were 22 guest. the host chooses 14 people to sit around a circular table and the others around another circular table. what is the probability that simon and brenna are not seated on the same table?

Answer 1

i've been sooo confused! i understand that for one table it's c(22,14) so that leaves the other one with c(8,8) possibilities? did i get that part right??

then i know there's a technique wherein total number of ways-total number of ways together=total number of ways apart but i can't seem to apply it to this! :(

Answer 2

@ParthKohli ? @sauravshakya ?

Answer 3

see, the total no. of ways in which 22 people can seat on two round tables in set of 14 and 8 are

13! * 7!

Now, the catch here is to find out the no. of ways in which simon and brenna can seat together:

If simon and brenna sit on table with 14 people or they can on the table with 8 people. The total no. of ways in which they can sit on same table is,

13! +7!

So, the probability = (13! +7!) / (13! * 7!)..

Answer 4

@bhaweshwebmaster don't we need combination in order to choose 14 out of the 22 and all that?

Answer 5

@Hero ? @TuringTest ? @saifoo.khan ?@Mertsj ? anyone?

Answer 6

@dydlf No Sir, we needn't use combination for this. This is the problem from "Basic Principles of Counting"... No need of combination...

Answer 7

but there are different ways of getting those 14 people out of the 22 :(

Answer 8

I think you are looking at wrong direction. factorial must do it...

Answer 9

@bhaweshwebmaster wait... why just factorial, then? because i could get a b c d e f g h i j k l m n o p q r s t u v -- 22 people and get a b c d e f g h i j k l m n as my 14 people or i j k l m n o p q r s t u v or even e f g h i j k l m n o p q r that's why i think combination is needed...
@amistre64 ? @Hero ? @ParthKohli ? anyone? what do you guys think?

Answer 10

Ya I understood what you were trying to say.. and probably you are right.. (or probably not).. I am 50-50... perhaps, math genuises on the forum could give us an insight better..

Answer 11

thanks @bhaweshwebmaster ! i'm really lost here. hopefully someone could help me! :(

Answer 12

im wondering how much of the information given is just for distraction

Answer 13

what is the probability that they are seated at the same table might be simpler to construct, then we can take the complement is one idea i have

Answer 14

there are 4 outcomes that i see; using table A and B and persons s and b, the possible seating outcomes are:

As, Ab
As, Bb
Bs, Ab
Bs, Bb

The possibility of you walking into the room and seeing s and b sitting at the same table would then be 2/4, and the probability that they are not seated at the same table is 2/4.

Answer 15

the possbility that s is at table A is say ... 1/14
the possbility that s is at table B is say ... 1/8

and same notions for person b if you wanna try to construct a tree chart, but im not sure what all you have been studying to know what method they are trying to get you to work out

Answer 16

another idea is: what is the probability of s being picked 1st, 2nd, .... 14th ? and b being picked 15th, 16th, ... 22nd?

Answer 17

P(s=1) = 1/22
P(s=2) = 1/21
P(s=3) = 1/20
...
\[P(s=1\to14)=\sum_{k=9}^{22}\frac{1}{k}\]
just wondering :)

Answer 18

Isnt it {2! 20C13} / {22C14}

Answer 19

@sauravshakya why 2! 20C13 for the numerator? can you explain please :D

Answer 20

Because simon and brenna are not seated on the same table

Answer 21

So, at first they must sit on different tables [ which mean 2!]
And we have to choose 13 people out of remaining 20 people to sit in the table [which gives 20C13]

Answer 22

So, 2! * 20C13 for the numerator

Answer 23

makes sense! thank you!!

Answer 24

welcome