document.write( "Question 1116837: A tray contains 15 chocolate chip cookies and 19 oatmeal raisin cookies. How many ways can a child select 4 cookies that include at least 1 chocolate chip cookie? \n" ); document.write( "
Algebra.Com's Answer #731763 by math_helper(2461)\"\" \"About 
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Assuming the cookies of each type are indistinguishable….
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\n" ); document.write( "\n" ); document.write( "Counting directly the number of possible ways of selecting 4 cookies with at least one or more chocolate chip cookies:
\n" ); document.write( " 1 choc chip + 3 oatmeal raisin
\n" ); document.write( " 2 choc chips + 2 oatmeal raisin
\n" ); document.write( " 3 choc chips + 1 oatmeal raisin
\n" ); document.write( " 4 choc chips + 0 oatmeal raisin
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\n" ); document.write( "\"+highlight%28+4%29+\" ways\r
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\n" ); document.write( "Using stars & bars….
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\n" ); document.write( "\n" ); document.write( "Again assuming indistinguishable cookies within each type, this problem maps to a stars and bars problem (with one adjustment) where there are two types (k=2) and we must select 4 of them (n=4).
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\n" ); document.write( "\n" ); document.write( "For k bars (types) and n stars (number of cookies), the arrangements are:
\n" ); document.write( "C(n+k-1, k-1) = C(n+k-1, n)
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\n" ); document.write( "\n" ); document.write( "Plugging in n=4, k=2: C(4+2-1, 1) = C(5,1) = 5
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\n" ); document.write( "\n" ); document.write( "The adjustment…
\n" ); document.write( "The above answer includes the case no chocolate chip cookies are selected, so we must subtract 1 for that case since it is a condition of the problem that at least one of that type is selected. That leaves us with 5-1 or \"+highlight%28+4+%29+\" ways.
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\n" ); document.write( "\n" ); document.write( "It may be tempting to use C(15,4)+C(15,3)+C(15,2)+C(15,1) or other counting formulas but those don't apply if the cookies of each type are indistinguishable.\r
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