document.write( "Question 4002: How many ways can you order a hamburger if you can order it with or without the following items: cheese, lettuce, tomatoes, pickles or onions? \n" ); document.write( "
Algebra.Com's Answer #1764 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
1. you can have 1 item only --> 5 possibilities\r
\n" ); document.write( "\n" ); document.write( "--> mathematically, this is worked out by 5!/1!4!\r
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\n" ); document.write( "\n" ); document.write( "2. you can have 2 items --> cheese with one of the other 4 items. lettuce with one of the other 3 items (not cheese, as that has been done), tomatoes with one of the other 2 items and then pickles with just onions. This totals 10 combinations\r
\n" ); document.write( "\n" ); document.write( "--> this is 5!/3!2!\r
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\n" ); document.write( "\n" ); document.write( "3. you can have 3 items:
\n" ); document.write( "ABC ACD ADE
\n" ); document.write( "ABD ACE
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\n" ); document.write( "\n" ); document.write( "BCD BDE
\n" ); document.write( "BCE\r
\n" ); document.write( "\n" ); document.write( "CDE\r
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\n" ); document.write( "\n" ); document.write( "--> Again totalling 10 --> 5!/3!2!\r
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\n" ); document.write( "\n" ); document.write( "4. you can have 4 items:\r
\n" ); document.write( "\n" ); document.write( "ABCD ACDE BCDE
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\n" ); document.write( "\n" ); document.write( "Totalling 5 combinations --> 5!/4!1!\r
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\n" ); document.write( "\n" ); document.write( "5. you can have 5 items --> total of 1 occurrence.\r
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\n" ); document.write( "\n" ); document.write( "So, total different ways is: 5 + 10 + 10 + 5 + 1 = 31\r
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\n" ); document.write( "\n" ); document.write( "jon
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