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PERMUTATIONS AND COMBINATIONS

PERMUTATIONS AND COMBINATIONS - PERMUTATIONS AND...

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PERMUTATIONS AND COMBINATIONS Topic 23, Section 2 - Combinations Section 1:        Permutations     Factorial representation of combinations Combination problems The sum of all combinations In permutations, the order is all important -- we count  abc  as different  from  bca .  But in  combinations  we are concerned only that  a, b,  and  have been  selected .   abc  and  bca  are the same combination.  Here are all the combinations of  abcd  taken three at a time:  abc    abd    acd    bcd . There are four such combinations.  We call this The number of combinations of 4 things taken 3 at a time. We will denote this number as   4 C 3 .  In general, n C k  =  The number of combinations of n things taken k at a time. Now, how are the number of combinations   n C k   related to the  number of  permutations n P k  ?  To be specific, how are the combinations    4 C 3   related to the permutations   4 P 3 Since the order does not matter in combinations, there are clearly  fewer combinations than permutations.  The combinations are  contained among the permutations -- they are a "subset" of the  permutations.  Each of those four  combinations , in fact, will give rise to  3! permutations:
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ab c   ab d   ac d   bcd acb   ad b   ad c   bdc bac   ba d   ca d   cbd bca   bd a   cd a   cdb cab   da b   da c   dbc cba   db a   dc a   dcb Each column is the 3! permutations of that combination.  But they are  all  one  combination -- because the order does not matter.  Hence there are  3! times as many  permutations as combinations.    4 C 3  , therefore, will be    4 P 3  divided by 3! -- the number of  permutations  that each combination  generates. 4 C 3   =    3!   =   4        3        2    1  2  3 Notice :   The numerator and denominator have the same number of  factors, 3, which is indicated by the lower index.  The numerator has 3  factors starting with the  upper  index and going down, while the  denominator is 3!.  In general,   n C k   =      k ! . n C k =   n    (    n       1)(   n       2)                     to k      
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factors                       k! Example 1.    How many combinations are there of 5 things taken 4 at a  time?     Solution .    5 C 4   =    5        4        3        2    1   2  3  4   =  5 Again, both the numerator and denominator have the  number of  factors  indicated by the lower index, which in this case is 4.  The numerator  has four factors beginning with the upper index 5 and going backwards.   The denominator is 4!.  Example 2.
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