Probability Question ..Need expert help

Let's take one instance of this arrangement

Alex Brian 3 4 5 6

With Alex fixed in position 1 and Brian in 2, rest of the people
can be arranged in 24 ways. Replace Alex with Chan. We will have
one more 24. Similarly replace Chan with Dan and then with Emily.
So, we have 4 * 24 = 96 ways.

Another approach to look at this will be to realize that other than
Brian and Fabian, rest 4 can always be arranged in 4! ways. Now,
how many such 4! ways depends on the position of Brian and Fabian.

One example:

_ B F _ _ _

You realize that _'s can be arranged in 4!. With Brian's position
fixed in 2, F can take 3 *more* positions (4, 5, and 6) towards it's
right (i.e F can take 3, 4, 5 & 6). So that is total 4 * 4! = 96

Similarly with Brian position fixed in 3

_ _ B F _ _

Here F can take total of 3 positions (4, 5 and 6). For each of those
positions, _'s can be arranged in 4! ways => total = 3 * 4! = 72.

I think you can work out the rest

HTH

voodoo_child wrote:

gmat_and_me wrote:
Brian in position 2 => 3, 4, 5, 6 can be arranged in 24 (4 * 3 * 2 * 1)
ways. But position 1 can change in 4 ways.

Hence total (4 * 24) = 96

GMAT_AND_ME = I didn't get why you are saying that "position 1 can change in 4 ways"..... Same goes with Brian occupying the second position. Can you please explain this?

Got it! Makes sense...Thanks

gmat_and_me wrote:Let's take one instance of this arrangement

Alex Brian 3 4 5 6

With Alex fixed in position 1 and Brian in 2, rest of the people
can be arranged in 24 ways. Replace Alex with Chan. We will have
one more 24. Similarly replace Chan with Dan and then with Emily.
So, we have 4 * 24 = 96 ways.

I think that 4*24 is incorrect because there are only 5 people, out of whom there is a condition on B and F. Essentially, you cannot have 4 options for Alex. IT will be only three.

E.g.

X B _ _ _ _ => I agree with 4P3 part. However, there are only three options for X....

Thoughts?

There are actually 6 people. If you see my first post, you can
ignore the condition that one person in absent, since empty
chair in one of the places is also an arrangement. For eg:

Alex Empty_Chair 3 4 5 6

is different from

Empty_Chair Alex 3 4 5 6

right ?

HTH

voodoo_child wrote:

gmat_and_me wrote:Let's take one instance of this arrangement

Alex Brian 3 4 5 6

With Alex fixed in position 1 and Brian in 2, rest of the people
can be arranged in 24 ways. Replace Alex with Chan. We will have
one more 24. Similarly replace Chan with Dan and then with Emily.
So, we have 4 * 24 = 96 ways.

I think that 4*24 is incorrect because there are only 5 people, out of whom there is a condition on B and F. Essentially, you cannot have 4 options for Alex. IT will be only three.

E.g.

X B _ _ _ _ => I agree with 4P3 part. However, there are only three options for X....

Thoughts?