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Jun 4 2007, 11:29 PM EDT
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Change: **this can be done with a 5 card hand. Just use the 3 card hand as a simpler model.** Example 2. Using exactly ‘r’ objects, with no identical objects. Problem - In how many ways can a 4 colors be drawn from a pack of 24 pencil crayons? Solution
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Jun 4 2007, 11:17 PM EDT
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Change: n=52;n=24; r=4 Way1-performed using a calculator Way 2-calculated by hand nCr = n! (n-r)!r! = 24C4 = 24!(24-4)! 4! = 10 626 ways = 24!20!4! = 10 626 waysEither way there are 10 626 ways of
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Jun 4 2007, 10:46 PM EDT
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Change: r=8r=4 Way1-performed using a calculator Way 2-calculated by hand nCr = n! (n-r)!r! = 24C4 = 24!(24-4)! 4! = 10 626 ways = 24!20!4! = 10 626 waysEither way there are 10 626 ways of drawing
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Jun 4 2007, 10:40 PM EDT
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Change: a king or queen4 cardspencils from a deckpack of 5224 cards.pencil crayons. Example 3. Not using exactly ‘r’ objects, but with some identical objects. Problem - How many sums of money can i make with 1 penny, 2 nickles and 1 dime . Solution
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Jun 4 2007, 10:17 PM EDT
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Change: a king or queen cards from a deck of 52 cards. Example 3. Not using exactly ‘r’ objects, but with some identical objects. Problem - How many different sums of money can be made using onei $5make bill,with two1 $10penny, 2 billsnickles and one
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May 25 2007, 2:52 PM EDT
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Change: The terminology has a great importance in getting an answer correct or incorrect. Pay attention to it. FormulaeFormula Combination notation nCr-->nCr--> = n! (see example 2 and flowchart) (n-r)!r! 2. Combinations in which some items are identical § (p+1) (q+1) (s+1) …-1 [to remove
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Jan 22 2007, 9:24 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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Nov 27 2006, 5:04 PM EST
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Change: one person on a committee. ReferencesCanton, Barbara J., Erdman, Wayne, Irvine, Jeff, Lim, Louis, Mclaren, Fran, Meisel, Roland W., Miller, David T., Speijer, Jacob. (2002). Mathematics of Data Management. Toronto: McGraw - Hill. (2006, July 7). Combinations Flowchart. Retrieved 2006, Nov 18 from http://www.gliffy.com/gliffy/#d=1107901&t=combinations
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Nov 21 2006, 9:40 PM EST
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Change: FormulaeFormula Combination notation nCr--> = n! (see example 2 and flowchart) (n-r)!r! Combinations in which some items are identical § (p+1) (q+1) (s+1) …-1 [to remove the null set] (see example 3) § (p+1) (q+1) (s+1) … [to include the null set] (see example 3) Not
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Nov 20 2006, 12:38 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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Nov 19 2006, 10:15 PM EST
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Change: NCONicoNCCOCoNoIgnore--> or 1. -Retain the null set when the question implies that no object at all can also be selected or states: up to, maximum of… - Remove the null set when the question
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Nov 19 2006, 10:14 PM EST
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Change: (p+1) (q+1) (s+1)(p+1)*(q+1)*(s+1) Step 3 -= (1+1)(2+1)(1+1)(1+1)*(2+1)*(1+1) = (2)(3)(2)(2)*(3)*(2) =12 Since a combination of money is required, the null set cannot be used. [ (p+1)(q+1)(s+1) …-1 ]Step 4 - =12-1 =11 Therefore 11 sums of money can be made. Example 4. Not
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Nov 19 2006, 10:08 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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Nov 19 2006, 10:07 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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Nov 19 2006, 10:03 PM EST
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Change: at least 1 spade. Example 2. Using exactly ‘r’ objects, with no identical objects. Problem - In how many ways can 4 cards be drawn from a deck of 52 cards? Solution - Recall from the flow chart, we use the formula nCr = n! (n-r)!r! n=52; r=4 Way1-performed
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Nov 19 2006, 9:50 PM EST
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Change: Solution-Step 1- Write out the cases. Case 1: Only one spade from 13 spades, four other cards from remaining 39 cards. 13C1 x 39C1 = 1 069 263 Case 2: Two spades from 13 spades, three other cards from remaining 39 cards. 13C2 x 39C3
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Nov 19 2006, 9:44 PM EST
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Change: Way 1 - preformed on a calculator.Way 2 - calculated by hand. Either way there are 270 725 ways of drawing 4 cards from a deck of 52 cards. Example 3. Not using exactly ‘r’ objects, but with some identical objects. Problem - How many
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Nov 19 2006, 9:32 PM EST
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Change: (ACP) 1287 + 27 885 + 211 926 + 712 842 + 1 069 263 = 2 023 203 Therefore there are 2 023 203 combinations possible of dealing a 5 card hand from 52 cards and deal out atleast 1 spade. Example 2. Using exactly
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Nov 19 2006, 9:30 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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Nov 19 2006, 9:02 PM EST
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Change: There were only format changes (bold, italics, etc.) in this version. See this version for details.
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