document.write( "Question 646779: Rework problem 9 from section 2.3 of your text. Assume that you have 6 dimes and 5 quarters (all distinct), and you select 4 coins. \r
\n" ); document.write( "\n" ); document.write( "(1) In how many ways can the selection be made? 330\r
\n" ); document.write( "\n" ); document.write( "(2) In how many ways can the selection be made if all the coins are dimes? 15\r
\n" ); document.write( "\n" ); document.write( "(3) In how many ways can the selection be made if you select 2 dimes and 2 quarters? 150\r
\n" ); document.write( "\n" ); document.write( "(4) In how many ways can the selection be made so that at least 2 coins are dimes? \r
\n" ); document.write( "\n" ); document.write( "I'm confused on number four. I tied using 6c2+6c3+6c4 and multiplying the same set but none of the answers are right. I know the problem can be set up as total-less than two options but i dont know how to set up the permutiation or combination. I do understand how to visualize the problem though \n" ); document.write( "

Algebra.Com's Answer #405946 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
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document.write( "Hi,
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document.write( "6 dimes and 5 quarters (all distinct), and you select 4 coins
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document.write( "(1) In how many ways can the selection be made? 11C4 =  330  Yes.
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document.write( "(2) In how many ways can the selection be made if all the coins are dimes? 6C4 = 15  Yes.
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document.write( "(3) In how many ways can the selection be made if you select 2 dimes and 2 quarters?  = 150 Yes.
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document.write( " (4) In how many ways can the selection be made so that at least 2 coins are dimes? 
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