probability of choosing numbers in ascending (descending) order

[Done]

this problem is giving me a hard time guys,


Manhattan GMAT’s football team has 99 players. Each player has a uniform number from 1 to 99 and no two players share the same number. When football practice ends, all the players run off the field one-by-one in a completely random manner. What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72, etc)?

(A) 1/64
(B) 1/48
(C) 1/36
(D) 1/24
(E) 1/16

(http://www.manhattangmat.com/)

how do we go about solving it?

[THUSSU]

if this question comes in the cat, skip it

[zango]

it will be 1/24
Here is how.. let us take any 4 numbers, say 10, 15, 20, 25 (just as an example, I have chosen these four numbers). What is true for these four numbers, is true for any other combination as well.

So, now, you can have 24 different arrangements of these numbers: P(4,4) = 4! = 24

10 15 20 25
10 15 25 20
10 25 15 20
10 25 20 15
10 20 15 25
10 20 25 15 = 6 combinations

20 10 15 25
20 10 25 15
20 15 10 25
20 15 25 10
20 25 10 15
20 25 15 10 = 6 combinations

15 10 20 25
..... = 6 combinations


25 10 15 20
25 10 20 15.
... = 6 combinations

Now, out of this, there is only 1 desired combination. So the probability = 1/24

It seems simple now - but I had to think over this for a couple of minutes - and would, in all probability - would have given it an attempt and then have skipped it - if it appeared in the paper. (I am assuming my solution is correct )

Cheers!

[Done]

nice explanation. It makes very much sense.