document.write( "Question 1018797: 18) In a box there are 8 red and 6 blue markers. How many ways can you select 3 markers if:
\n" ); document.write( "(a) exactly 1 red is selected?
\n" ); document.write( "(b) no more than 2 blue are selected?
\n" ); document.write( "(c) there are more blue than red selected?\r
\n" ); document.write( "\n" ); document.write( "I'm not sure where to begin can you please guide me through?
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Algebra.Com's Answer #634869 by Boreal(15235)\"\" \"About 
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8 ways to select one red.
\n" ); document.write( "6C2 ways to select 2 blue
\n" ); document.write( "that is 15.
\n" ); document.write( "numbering the blues 1 through 6
\n" ); document.write( "12
\n" ); document.write( "13
\n" ); document.write( "14
\n" ); document.write( "15
\n" ); document.write( "16
\n" ); document.write( "23
\n" ); document.write( "24
\n" ); document.write( "25
\n" ); document.write( "26
\n" ); document.write( "34
\n" ); document.write( "35
\n" ); document.write( "36
\n" ); document.write( "45
\n" ); document.write( "46
\n" ); document.write( "56
\n" ); document.write( "So the number of ways are 8*15=120
\n" ); document.write( "no more than two blue mean 15 ways for 2+6 ways for 1+the number of ways one can select 3 from 8 red (0 blue). That is 8C3 or 56 ways, so the total is 15+6+56=77 ways.
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\n" ); document.write( "more blue than red.
\n" ); document.write( "3 blue and 0 red. That is 6C3=20 ways. 6C3 is the number of ways one can select 3 items from 6. It si 6!/3!/3!, or 6*5*4*3!/3!3!
\n" ); document.write( "2 blue and 1 red. That is 6C2*8=120 ways and the total is 140 ways.
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