# Re: consecutive numbers

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Posted by Brad Paul on September 22, 2002 at 12:27:22:

In Reply to: consecutive numbers posted by oldguy on September 21, 2002 at 15:25:26:

: I meant any consecutive order i.e. 1 5 21 23 25. 6 11 13 24 25. ect.

If the question is:

How many different ways can one arrange m numbers such that they are in
increasing order when picking from a set of n unique numbers, where n>m.

Then I agree with Soroban:

N=n!/(m! (n-m)!)

because the previous equation gives how many combinations of m are
there when selecting without replacement from a set of n when the order
of m does not matter. Also for every set of m there are m! ways or
arranging that set. However, only one of those ways corresponds to
increasing order and every set of m can be arranged into increasing
order.

If the question is:

If one picks m numbers from a set of n unique numbers what is the
probability that the set of m is in increasing order? This time order
does matter which means the number of possible sets of m numbers pulled
form n is:

No=n!/(n-m)!

and of those sets only

N=n!/(m! (n-m)!)

are in increasing order. Therefore the probability the set pick is in
increasing order is:

P=N/No=1/m!

Another way to word this is in two steps:

1) I'm going to pick m numbers from n numbers. What is the probability
that I get m numbers? (This is one of those too obvious questions)

p1=1;

2) What is the probability they are in order?

p2=1/m!

P=p1 * p2

PS

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