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Unformatted text preview: MATH 135 Winter 2009 Lecture VII Notes Binomial Coefficients We want to determine a general expression for ( a + b ) n where n is a positive integer. This would save us from expanding this product every time. (Does this concept sound familiar?) The expansions for n = 1 , 2 , 3 may be familiar: ( a + b ) 1 = a + b ( a + b ) 2 = a 2 + 2 ab + b 2 ( a + b ) 3 = ( a + b ) 2 ( a + b ) = ( a 2 + 2 ab + b 2 )( a + b ) = a 3 + 3 a 2 b + 3 ab 2 + b 3 ( a + b ) 4 = ( a + b ) 2 ( a + b ) 2 = ( a 2 + 2 ab + b 2 )( a 2 + 2 ab + b 2 ) = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 We can see a pattern relatively quickly in how the powers of a and b enter into these expansions, but the coefficients are more tricky. To see how this works, we need to look first at permutations and combinations. Question n students wish to form a single line. How many different possible lines are there? Answer: n ! = n ( n- 1)( n- 2) (3)(2)(1) (There are n possible students for the first position, then n- 1 for the second position, then...
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