# The number of ways to draw X items from a set of n items (without replacement).

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# [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

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• ### a "discard" pile* of n-X 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.

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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

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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

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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

DCthese 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

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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

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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!/(4-2)! = 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

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If order doesn't matter in the selected pile--for example, when we consider AC BD to be the same as CA BD--then 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.)

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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