# Probability

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• Jan 22nd 2007, 01:12 PM
mauro21pl
Probability
Hi to all
I have few short questions if you would like to look at them
a)in how many ways can 3 boys and 3 girls sit in a row?
b)in how many ways can 3 boys and 3 girls sit in the row if the boys and the girls are each to sit together
c)in how many ways if only the boys must sit together?
d)in how many ways if no two people of the same sex are allowed top sit together?
Thank you
• Jan 22nd 2007, 03:22 PM
Plato
What have you tried on any of these?
Have you done any work on them?

The answer to part c is: 2(3!)(3!).
• Jan 22nd 2007, 03:57 PM
mauro21pl
thanks
Yes i done part a b and d.I just needed answere for c. I wanted just check the answeres.Thanks
• Jan 22nd 2007, 04:00 PM
mauro21pl
Actually i've got defferent answere for c which probably is wrong but i've got the same answere what you have for part b
• Jan 22nd 2007, 04:10 PM
Plato
Actually parts b & c have the same numerical answer.
The reasons for that are entirely different.
You may be reading “if only” as “only if”. They are not the same!

In the future, if you post your answers we will know that you are not just on a fishing expedition to see if someone will do your work for you.
• Jan 23rd 2007, 09:15 AM
Soroban
Hello, mauro21pl!

Quote:

a) In how many ways can 3 boys and 3 girls sit in a row?
With six people, there are: . $6! = 720$ ways.

Quote:

b) In how many ways can 3 boys and 3 girls sit in the row
if the boys and the girls are each to sit together?

There are two arrangements of the genders: . $BBBGGG$ or $GGGBBB$

In each, the three boys can be ordered in $3!$ ways
. . and the girls can be ordered in $3!$ ways.

Therefore, there are: . $2 \times 3! \times 3! \:=\;72$ ways.

Quote:

c) In how many ways if only the boys must sit together?
Duct-tape the boys together.
Then there are four "people" to arrange. .There are $4!$ arrangements.

For each arrangement, the three boys can be ordered in $3!$ ways.

Therefore, there are: . $4! \times 3! \:=\:144$ ways.

Quote:

d) In how many ways if no two people of the same sex are allowed to sit together?
Since the boys and girls must alternate,
. . there are two arrangements: . $BGBGBG$ or $GBGBGB$

For each arrangement, the three boys can be ordered in $3!$ ways
. . and the three girls can be ordered in $3!$ ways.

Therefore, there are: . $2 \times 3! \times 3! \:=\:72$ ways.