The number of ways to draw X items from a set of n items (without replacement).
(Click on the right arrow to begin.)
First we start with a mother set, say,
[A B C D]
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There are 4! = 24 permutations of the set, [A B C D]. So our count starts at 24.
ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACBD BCAD CBAD DBAC
ACDB BCDA CBDA DBCA
ADBC BDAC CDAB DCAB
ADCB BDCA CDBA DCBA
(Click right arrow to continue.)
Now, the act of drawing X items is tantamount to splitting that initial permutation of n items into two piles:

a "selected" pile* of X items, and

a "discard" pile* of nX items.
* I often call these "success" and "failure" piles in class. Successes are the things you're interested in (looking for) and failures are those things you're not interested in.
(Click the right arrow to continue.)
AB CD BA CD CA BD DA BC
AB DC BA DC CA DB DA CB
AC BD BC AD CB AD DB AC
AC DB BC DA CB DA DB CA
AD BC BD AC CD AB DC AB
AD CB BD CA CD BA DC BA
(Click right arrow to continue.)
So let's split each permutation into a selected (success) pile with X = 2 members and a discard (failure) pile with n  X = 2 members.
split
selected discarded
AB CD BA CD CA BD DA BC
AB DC BA DC CA DB DA CB
AC BD BC AD CB AD DB AC
AC DB BC DA CB DA DB CA
AD BC BD AC CD AB DC AB
AD CB BD CA CD BA DC BA
(Click right arrow to continue.)
In practice, the order of the members (items) in the discard pile never matters; it's always a set. So, for example, the two permutations AB CD and AB DC count as one arrangement.
Because CD is the same as
DC, these two count as one.
AB CD BA CD CA BD DA BC
AB DC BA DC CA DB DA CB
AC BD BC AD CB AD DB AC
AC DB BC DA CB DA DB CA
AD BC BD AC CD AB DC AB
AD CB BD CA CD BA DC BA
(Click right arrow to continue.)
We can reduce the total count of n! = 24 by the number of ways we can arrange the discard pile, (n  X)! = 2! = 2. Our count is now 4!/2! = 12 unique arrangements.
AB CD BA CD CA BD DA BC
AC BD BC AD CB AD DB AC
AD BC BD AC CD AB DC AB
(Click right arrow to continue.)
This then is the number of ways to draw X = 2 items from a set of n = 4 when order does not matter in the discard pile and order does matter in the selected pile: 4!/(42)! = 12. That is, we draw out one list and one set, which is a partial permutation.
AB CD BA CD CA BD DA BC
AC BD BC AD CB AD DB AC
AD BC BD AC CD AB DC AB
(Click right arrow to continue.)
If order doesn't matter in the selected pilefor example, when we consider AC BD to be the same as CA BDthen we further reduce this permutation by X! = 2! = 2. So now 4!/(2!2!) = 6. We draw out two sets. This is a combination.
Example 1
Order matters in the selected pile.
Order doesn't matter in the discard pile.
 Five finalists in a drawing, n = 5.
 Prizes of $50,000 and $10,000 awarded to two of the finalists, X = 2. Order matters for the selected pile, because getting $50,000 isn't the same as $10,000.
 In how many different ways can we get two prize winners out of five finalists?
(Click the down arrow for the solution.)
Example 1 Solution
Order matters in the selected pile.
Order doesn't matter in the discard pile.
 The number of permutations, 5! = 120.
 Because the order of the members in the discard (or losing) pile doesn't matter, we reduce this permutation by (n  X)!:
 5!/3! = 120/6 = 20, a partial permutation.
AB AC AD AE BA BC BD BE CA CB CD CE
DA DB DC DE EA EB EC ED
(Click on the right arrow for the next example.)
Example 2
Order doesn't matter in the selected pile.
Order doesn't matter in the discard pile.
 There is a committee of 5 people. There needs to be 3 people at the meeting to have a quorum (to be able to vote on things).
 In how many ways, out of 5 members, can exactly 3 members show up to the meeting?
(Click the down arrow for the solution.)
Example 2 Solution
Order doesn't matter in the selected pile.
Order doesn't matter in the discard pile.
 The number of permutations is 5! = 120.
 Because order doesn't matter for either the selected pile or the discard pile, we reduce 5! by both X! = 3! and (n  X)! = 2!:
 5!/(3!2!) = 120/12 = 10, a combination.
ABC ABD ABE ACD ACE ADE
BCD BCE BDE CDE
(That's it. Close the window when you're done.)
1. Probability  Drawing Without Replacement
By smilinjoe
1. Probability  Drawing Without Replacement
The basics on how to count the number of ways to draw k items from a mother set of n items.
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